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https://docs.google.com/document/d/16Kj6rng-45JDKIVWRLsMajUAri0fhdy-/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true CURRENT ELECTRICITY CURRENT ELECTRICITY Flow of electric charge constitutes electric current. For a given conductor, if 'δQ' charge flows through a cross-section of area A in time 'δt', then the average electric current through the conductor is given as I = and its instantaneous value is . MECHANISM OF CURRENT FLOW IN METALLIC CONDUCTOR When an external potential difference is applied across a metallic conductor then an electric field is set up within the conductor. Applied electric field → Force on electrons → drift of electrons Due to the externally applied electric field electrons drift with an average velocity called drift velocity. This causes an electric current Total charge crossing a cross-section in one second is equal to I = neAvd. Here Avd is the volume of a cylinder of cross-section A length vd and ne is charge density of charge carriers (e.g. electrons). The current density is defined by J = Ι/A Example1 : A steady current passes through a cylindrical conductor. Is there an electric field inside the conductor ? Solution : Yes; No doubt under steady state conditions in electrostatics when a conductor is charged, electric field inside it is zero as metal is an equipotential surface. However when a potential difference is applied across a conductor and a steady current flows though it, the condition no longer remains static and there exists an electric field inside the conductor. OHM’S LAW It states that "the potential difference across a conductor is directly proportional to the current flowing through it at a given temperature". ⇒ At constant temperature the constant 'R' is called resistance of the conductor. Resistivity (ρ) and conductivity (σ): The resistance R of a given conductor is directly proptinal to length () and inversitional proptional cross-sectional area (A) such that R = ρ , where ρ = resistivity of the material of the given conductor. Its S.I. unit is Ω m. Reciprocal of resistivity is called the electrical conductivity (σ) of the material, thus σ = = whereas reciprocal of resistance is called conductance of the given conductor. S.I. unit of conductivity σ is (Ω - m)-1 and is usually written as mho/m. Temperature Dependence of Resistivity: The conductivity of a metal decreases as its temperature is increased. Thus resistivity ρ increases with the rise in temperature. If ρT and ρ0 represent the resistivities at temperatures T and T0 respectively, then for small temperature variations, ρT= ρ0 [1 + α(T − T0)] Where α is called the temperature coefficient of resistivity. The resistivity varies over a very wide range. For metals (good conductor) ρ ≈ 10-8 Ω-m and for insulators ρ ≈ 1017 Ω-m Semiconductors (silicon, germanium ) have intermediate value much smaller than insulator but much larger than metals. Temperature coefficient of resistivity is negative for semiconductors and positive for the metals. For superconductors resistivity is zero. Thermistor: A thermistor is a semiconductor electronic device in which the resistance decreases as its temperature increases. This is used as a thermometer. The temperature coefficient of resistivity is negative for semiconductors, hence thermistors are usually prepared from oxides of various metals such as nickel, iron, cobalt and copper etc. A thermistor is used to detect small changes in temperature of the order of even 10-3 0C. Colour code for carbon Resistors: The four bands indicate digit -1, digit-2, multiplier and tolerance respectively and the values of different colours are given in the following table. Resistance code (in Ω) Colour Digit Multiplier Tolerance Black 0 1 Brown 1 10 Red 2 102 Orange 3 103 Yellow 4 104 Green 5 105 Blue 6 106 Violet 7 107 Gray 8 108 White 9 109 Gold 0.1 5% Silver 0.01 10% Sometimes the carbon resistor indicates only three bands and the tolerance is missing from the colour code. This means tolerance has to be taken as 20%. Example2: Find the resistance of a carbon resistor if the colour code from left to right indicates brown, yellow, green and gold. Solution: Use diagram ↓ ↓ ↓ ↓ 1 4 × 105 ± 5% R = (14 × 105 ± 5%) Ω = (1.4 × 106 + 0.07 × 106 ) Ω = (1.4 ± 0.07)MΩ KIRCHHOFF’S LAWS Junction Rule: It is based on the law of conservation of charge. At a junction in a circuit the sum of incoming currents is always equal to the sum of outgoing currents. In otherwords the algebraic sum of the currents at a junction is zero. Loop rule The algebraic sum of the changes in potential around any closed path is zero. It is based on the law of conservation of energy. ∙ In case of a resistor of resistance 'R' potential will decrease in the direction of current. Hence, for the shown conductor Va – Vb = IR ∙ For an emf source, the potential changes will be obtained as illustrated below, Emf = ε, internal resistance = r Va – Vb = ε + ir Emf = ε, internal resistance = r Va – Vb = −ε + ir Students can use any sign convention which they find easy. Example3: In the series circuit shown, E,F,G,H are cells of emf 2V,1V,3V and 1V respectively, and their internal resistance are 2, 1, 3 and 1Ω respectively. Calculate (i) the potential difference between B and D and (ii) the potential difference across the terminals of each of the cells G and H. Solution: Let us redraw the circuit. At junction D, we have applied the junction rule, whereby we get current in DB as shown. Loop BADB 2I1 - 2 + 1 + I1 + 2 (I1 − I2) = 0 ⇒ 5I1 − 2I2 = 1 Loop DCBD −3+3I2+I2+1−2(I1−I2)=0 ⇒ 6I2 – 2I1 = 2 (i) VBD = 2(I1) − 2 + 1 + I1= 3 I1 − 1 = (ii) Terminal voltage of G = |−3 + 3.I2| = = Terminal voltage of H = . GROUPING OF RESISTANCES Resistance in series Let the equivalent resistance between A & B equals Req , by definition. Req = . . . (1) Using Kirchoff's 2nd rule for the loop shown in figure, V = IR1 + IR2 + IR3 . . . (2) From (1) and (2) Req = R1 + R2 + R3 Resistance in parallel Here again, Req = . . . (1) I = i1 + i2 + i3 = . . . (2) From (1) and (2) Example 4: Find the equivalent resistance between A and B in the circuit shown here. Every resistance shown here has a magnitude of 2 Ω. Solution: Points C, O & D are at the same potential. Therefore, resistances AO, AC and AD are in parallel . Similarly BC, BO and BD are in parallel. Similarly BC, BO and BD are in parallel. ∴ RAB = ×(2Ω) + × (2Ω) = Ω = 1.33 Ω. ENERGY, POWER AND HEATING EFFECT When a current I flows for time t from a source of emf E, then the amount of charge that flows in time t is Q = It. Electrical energy delivered W = Q.V = VIt Thus, Power given to the circuit, = W/t = VI or V2/R or I2R In the circuit E. I = I2R + I2 r, where EI is the rate at which chemical energy is converted to electrical energy, I2R is power supplied to the external resistance R and I2r is the power dissipated in the internal resistance of the battery. An electrical current flowing through conductor produces heat in it. This is known as Joule's effect. The heat developed is given by H = I2.R.t joule, where I = current in ampere , R = resistance in Ω, t = time in second. Maximum Power Theorem In a circuit, for what value of the external resistance the maximum power be drawn from a battery? For the shown network power developed in resistance R equals ( I = and P = I2R ) Now, for dP/dR = 0 (for P to be maximum ) ⇒ R + r = 2R ⇒ R = r ⇒ The power output is maximum, when the external resistance equals the internal resistance. R = r Example5 : A copper wire having a cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at 25oC and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire (a) Find the time in which the wire starts melting. The change of resistance of the wire with temperature may be neglected. (b) What will this time be, if the length of the wire is doubled? Density of Cu = 9 × 103 Kg m-3 specific heat of Cu = 9 × 10-2 Cal Kg-1 oC-1, M.P. (Cu) 1075 oC and specific resistance = 1.6 × 10-8Ωm. Solution : (a) Mass of Cu = Volume × density = 0.5 × 10-6 × 0.1 × 9 × 103 = 45 × 10-5 Kg. Rise in temperature = θ = 1075-25 = 1050 oC. Specific heat = 9 × 10-2 Kg-1 oC × 4.2 J ⇒ I2Rt = mSθ (b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, wire will start melting in the same time. WHEATSTONE BRIDGE For a certain adjustment of Q, VBD = 0, then no current flows through the galvanometer. ⇒ VB = VD or VAB= VAD ⇒ I1.P = I2.R Likewise, VBC = VDC ⇒ I1.Q = I2.S Dividing, we get, GROUPING OF IDENTICAL CELLS 1. Series Grouping emf of the cell is ε and internal resistance is r. R is the external resistance, I is the current passing through the circuit and n is the total number of cells. Applying Kirchhoff’s law ε − ir + ε − ir + ........ (to n times ) − iR = 0 ⇒ 2. Parallel grouping Applying Kirchhoff’s law ⇒ 3. Mixed grouping Number of rows is m and number of cells in each row is n Apllying kirchhoff’s law ⇒ I = For current through R to be maximum, mR = nr ⇒ R = = one line total internal resistance/No. of Line RC-CIRCUIT Charging: Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0. Now, 'C' is getting charged. If the charge on capacitor at time 't' is q. writing the loop rule, + IR − E = 0 Integrating ⇒ At t = 0, q = 0 and at t = ∞, q = E C (the maximum charge.) = qmax Thus, Discharging Consider the same arrangement as we had in the previous case with one difference that the capacitor has charge qo for t<0 and the switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t then the loop equation is given as Flip the switch to 2 Integrating, at t = 0, q = q0 t = t, q = q '-ve' sign indicates that the discharging current flows in a direction opposite to the charging current. Example6: Calculate the steady-state current in the 2Ω resistor shown. The internal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 μF. Solution : The resistance of the parallel combination of 2Ω and 3Ω resistors is given by This resistance is in series with 2.8 Ω giving a total effective resistance = 1.2 + 2.8 Ω = 4 Ω. In the steady state, charge on the capacitor C has stablised and hence no current passes through 4Ω resistor which is in series with the capacitor. Thus the current through the circuit = 6/4 = 1.5 A, VAB = 1.5 × 1.2 = 1.8 V, I through 2Ω resistor = 1.8/2 = 0.9 A. AMMETER An Ammeter is an instrument used for measuring current in electrical circuits. A galvanometer is a low resistance instrument. A large current passing through it may damage the instrument Changing the range of an ammeter Suppose the ammeter gives full scale deflection when a current Ig flows through it. Now if we want to convert the reading of the ammeter in such a manner that it gives full scale deflection for a higher current I in the branch of the circuit, we connect a small resistance S in parallel to the coil of the galvanometer, which has a resistance G. The resistance value is so chosen that out of the total current I only Ig flows through the coil and the remaining current flows through S. As potential difference across S = potential difference across G. ⇒ (I – Ig)S = IgG ⇒ S = So, effectively this ammeter will measure current up to I ampere and its effective resistance = . In practice G is large as compared to S. Therefore the effective resistance of the ammeter equals RA ≈ S, which is small. An ideal ammeter has zero resistance. VOLTMETER A voltmeter is an instrument used for measuring potential difference across the two ends of a current carrying conductor. It is connected in parallel with the conductor across which the potential difference is to be measured. The current through the conductor should not change on connecting the voltmeter, and so the voltmeter should draw a very small current, i.e. its resistance has to be high. When a galvanometer is used to measure potential difference across the ends of a current carrying conductor, a high resistance R is connected in series with the galvanometer. Consider the diagram shown. Suppose the galvanometer gives full scale deflection when a current Ig passes through its coil. If G is resistance of the galvanometer coil then: VOLTMETER Potential difference to be measured = V = ⇒ R = - G So, effectively the voltmeter has resistance = Rv = (G + R) In practice Rv is very large compared to G. An ideal voltmeter should possess infinite resistance. POTENTIOMETER A potentiometer is used to compare electromotive forces of two cells or to measure the internal resistance of a cell. Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform cross sectional area, the potential drop across any portion of the wire is directly proportional to its length. Consider the network shown in the diagram : Here, Now, let us consider that an EMF source having EMF same as (VA – VB) calculated above and internal resistance ‘r’ connected in parallel to R1. Suppose that a current ‘I2’ passes through the EMF source We have, applying Kirchoff’s rules, the following equations R2 (I1+I2) + R1 I1 + R (I1+ I2) = E . . . (i) R1 I1 = I2 r + E′ . . . (ii) Solving (i) & (ii) we get I2 = For the chosen value of E' that is we get I2 = 0 Thus, taking a length of uniform resistance wire between A and C instead of the two resistors R1 & R2, it is possible to get zero (null) deflection in the galvanometer in the circuit shown below, provided that. E1 < IRAC Let 1 be the length AB of the wire, corresponding to zero deflection in galvanometer, and let B be the point where the movable pointer ‘P’ (also called Jockey) makes contact with wire AC. If the same experiment is repeated but with another cell of EMF E2, we will be getting a different length ‘2’ at the instant galvanometer shows null deflection. If in both the cases the rheostat ‘R’ is kept unchanged, . The arrangement shown in figure (iii) is called Potentiometer and we use it to measure EMF using the above relationship. Application : A potentiometer can be used to compare emfs of two cells or to measure internal resistance of a cell. The method to compare the emfs of two cells has already been explained. When we want to measure the internal resistance of a cell, consider the following formula : r = , where r = internal resistance of the cell, E = EMF, V = potential difference across the cell during current flow, S = resistance of resistance box, 1 , 2 are two balancing lengths. Example7: A potentiometer wire of length 100 cm has a total resistance of 10Ω. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A cell of emf 10 mV is balanced against a length of 40cm of potentiometer wire. What is the value of the external resistance R ? Solution : As shown in the figure, if R is the unknown resistance, the current in the circuit I = Now as the 100 cm wire has a resistance of 10 Ω, the resistance of 40 cm of wire will be 40 × (10/100) = 4 ohm. Potential drop across 40 cm wire will be V = I ×4 but here V = 10 mv (given) Hence, = i.e. R = 790 Ω. OBJECTIVE 1: A copper wire of diameter 1.02 mm carries a current of 1.7 amp. Find the drift velocity (vd) of electrons in the wire. Given n, number density of electrons in copper = 8.5 × 1028 /m3. (A) 1.75 mm/sec (B) 1.25 mm/sec (C) 2.5 mm/sec (D) 1.5 mm/sec Ans. (d) Solution: I = 1.7 A J = current density = = nevd = 8.5 × 1028 × (1.6 × 10-19 ) × vd ∴ vd = = 1.5 × 10-3 m/sec. = 1.5 mm/sec. 2: A cylindrical conductor of length  and inner radius R1 and outer radius R2 has specific resistance ρ. A cell of emf ε is connected across the two lateral faces of the conductor. Find the current drown from the cell. (A) I = (B) I = (C) I = (D) I = Ans. (a) Solution: Consider the differential element of the cylinder as shown in the figure. ⇒ R = I = ⇒ I = 3: A copper wire of resistance 4Ω is melted and redrawn to thrice its original length. Find the resistance of stretched wire. (A) 40 Ω (B) 60 Ω (C) 45 Ω (D) 36 Ω Ans. (d) Solution : Since volume of the wire does not change. ⇒ 1A1 = 2 A2 where 1 and A1 are the initial length and cross-section of the wire, and 2 and A2 are the final length and cross-section. ⇒ . . . (1) ∴ R1 = ρ and R2 = ρ ⇒ . . . (2) ⇒ as 2 = 31 ⇒ R2 = 9R1 = (9 × 4)Ω = 36 Ω 4: An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which bulb will glow more? (A) (B) (C) (D) Ans. (b) Solution : ∴ resistance of first bulb is and resistance of the second bulb is In series same current will pass through each bulb ∴ Power developed across first is and that across second is ⇒ as ⇒ ⇒ ⇒ The bulb rated 220 V & 40 W will glow more. 5: A battery of emf 1.4 V and internal resistance 2Ω is connected to a 100 Ω resistor through an ammeter. The resistance of the ammeter is 4/3 Ω. A voltmeter is also connected to find the potential difference across the resistor. The ammeter reads 0.02A. What is the resistance of the voltmeter? (A) 200 (B) 300 (C) 400 (D) 150 Ans. (a) 6: A battery of emf 1.4 V and internal resistance 2Ω is connected to a 100 Ω resistor through an ammeter. The resistance of the ammeter is 4/3 Ω. A voltmeter is also connected to find the potential difference across the resistor. (iii) The voltmeter reads 1.10 V. What is the error in reading? (A) 0. 3 V (B) 0.43 V (C) 0.53 V (D) 0.23 V Ans. (d) Solution: 5 & 6 (ii) Rv = 200 (iii) Potential difference across the voltmeter = =1.33 V. Voltmeter reading = 1.10 V Error = 1.33 - 1.10 = 0.23 V. 7: Five equal resistances each of value R are connected to form a network as shown in figure. Calculate the equivalent resistance of the network between the points B and D. (A) (B) (C) (D) Ans. (c) Solution : The circuit can be redrawn as in the figure. (a) for points B and D, resistance (P+R), G an (Q+S) are in parallel. With P = Q = R = S = G = R So, i.e. 8: Five equal resistances each of value R are connected to form a network as shown in figure. Calculate the equivalent resistance of the network between the points A and C, and A (A) R (B) R/2 (C) 2 R (D) 5 R Ans. (a) Solution : for points A and C, the given network is balanced wheatstone bridge as . So excluding G, (P+Q) is in parallel with (R+S) i.e. i.e. (Req)AC = R 9: Five equal resistances each of value R are connected to form a network as shown in figure. Calculate the equivalent resistance of the network between the points A and C (A) (B) (C) (D) Ans. (a) Solution : for points A and B, starting from opposite side of AB, (Q+S) in parallel with G gives RBCD = RBCD in series with R gives RBDO = And RBDO in parallel with P gives the equivalent resistance between AB, i.e. (Req)AB = i.e. (Req)AB = 10: Figure shows a potentiometer circuit for determining the internal resistance of a cell. When switch S is open, the balance point is found to be at 76.3 cm of the wire. When switch S is closed and the value of R is 4.0 Ω, the balance point shifts to 60.0 cm. Find the internal resistance of cell E′. (A) (B) (C) (D) Ans. (b) Solution : , But , hence 11: In the circuit shown in figure each cell has emf 5V and has an internal resistance of 0.2 ohm. What is the reading of ideal voltmeter V. (A) 0V (B) 1V (C) 2V (D) 5V Ans. (a) Solution: As internal resistance of an ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as ⇒ r' = r Now as the given 8 batteries are discharging in series i.e. Eeq = 8 × 5 = 40 V and req = 8 × 0.2 = 1.6 Ω so current in the circuit I = = 25 A Hence potential difference across the required battery v = E – Ir = 5 – 25 × 0.2 = 0V. 12: When two resistances X and Y are put in the left hand and right hand gaps in a wheatstone meter bridge, the null point is at 60cm. If X is shunted by a resistance equal to half of itself then find the shift in the null point. (A) 26.7 cm (B) 36.7 cm (C) 46.7 cm (D) 96.7 cm Ans. (a) Solution: Arrangement is shown in the figure. . . . . (1) When X is shunted then resistance in the left gap becomes . . . (2) Now ∴ Shift = 60 – 33.3 = 26.7 cm. 13 : Each resistor in the network of the given figure has a resistance of 10Ω. The resistance between points A and B is (A) 10 Ω (B) 20Ω (C) 30Ω (D) 40 Ω Solution: Resistors in arms CD and EF are in series which add up to 10+10=20 Ω. Resistors in arms CE and EF are also in series which add upto 20 Ω Hence resistance between points C and F, is given by or R = 10 Ω Hence, resistance between A and B = 10 Ω + 10 Ω + 10 Ω = 30 Ω ∴ (C) 14: In a gas discharge tube if 3 × 1018 electrons are flowing per sec from left to right and 2 × 1018 protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section (A) 0.48, left to right (B) 0.48 A, right to left (C) 0.80A, left to right (D) 0.80 A, right to left Solution : As current is rate of flow of charge in the direction in which positive charge will move, the current due to electron will be ie = = 3 × 1018 × 1.6 × 10−19 = 0.48 A (Opposite to the motion of electrons, i.e. right to left) Current due to protons ip = = 2 × 1018 × 1.6 × 10−19 = 0.32 A (Right to left) so total I = ie + ip = 0.48 + 0.32 = 0.80 A (Right to left) Hence correct answer is (D) 15 : The current in a wire varies with time according to the equation I = 4 + 2t, where I is in ampere and t is in sec. The quantity of charge which has passed through a cross-section of the wire during the time t = 2 sec to t = 6 sec will be (A) 60 coulomb (B) 24coulomb (C) 48 coulomb (D) 30 coulomb Solution : Let dq be the charge which has passed in a small interval of time dt, then dq = idt = (4 + 2t)dt Hence total charge passed between interval t = 2 sec and t = 6 sec q = = 48 coulomb ∴ (C) 16 : A uniform copper wire of length 1 m and cross-sectional area carries a current of 1 A. Assuming that there are free electrons per m3 in copper, how long will an electron take to drift from one end of the wire to the other? (A) sec (B)sec (C) sec (D) sec Solution : The drift velocity of electrons is given by If  is the length of the wire, the time taken is =sec ∴ (D) 17 : A electric current of 16 A exists in a metal wire of cross section 10−6 m2 and length 1m. Assuming one free electrons per atom, the drift speed of the free electrons in the wire will be (Density of metal =5 × 103 kg/m3, atomic weight = 60) (A) 5 × 10−3 m/s (B) 2 × 10−3 m/s (C) 4 × 10−3 m/s (D) 7.5 × 10−3 m/s Solution : According to Avogadro's hypothesis so n = NA Hence total number of atoms n = = 5 × 1028 /m3 As I = ne eA vd Hence drift velocity vd = vd = = 2 × 10−3 m/s ∴ (B) 18: A copper wire is stretched to make it 0.1 % longer. The percentage change in its resistance is (A) 0.2 % increase (B) 0.2% decrease (C) 0.1 % increase (D) 0.1 % decrease Solution : For a given wire, R = with L × s = volume = V = constant so that R = ρ = 2 = 2 (0.1 %) = 0.2 % (increase) ∴ (A) 19 : The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5Ω and is 1m long. The resistance which must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is (A) 1985 Ω (B) 1990 Ω (C) 1995 Ω (D) 2000 Ω Solution : In order to have a potential drop of 5 mV = across a wire of resistance 5 Ω, the current flowing in the wire should be If R is the resistance to be connected in series with the wire, then which gives R = 1995 Ω ∴ (C) 20 : A battery of 10 volt is connected to a resistance of 20 ohm through a variable resistance R. The amount of charge which has passed in the circuit in 4 minutes, if the variable resistance R is increased at the rate of 5 ohm/min. (A) 120 coulomb (B) 120 loge2 coulomb (C) coulomb (D) coulomb Solution : I= dq = 12 V q = 12 V = 12 V (loge 40 – loge20) = 12 × 10 × loge2 ∴ (B) 21: In the given circuit, R1 = 10 Ω, R2 =6 Ω and E = 10 V. Then effective resistance (A) Effective resistance of the circuit is 20Ω (B) Effective resistance of the circuit is 30Ω (C) Effective resistance of the circuit is 40Ω (D) Effective resistance of the circuit is 50Ω Solution: Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be Effective resistance of the circuit Reff = = 20 Ω current through the circuit I = = (1/2) amp Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. ∴ (A) 22: In the given circuit, R1 = 10 Ω, R2 =6 Ω and E = 10 V. Then reading of A1 (A) Reading of A1 is 1 amp. (B) Reading of A1 is 1/2 amp. (C) Reading of A1 is 2 amp. (D) Reading of A1 is 3 amp. Solution: Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be Effective resistance of the circuit Reff = = 20 Ω current through the circuit I = = (1/2) amp Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. ∴ (B) 23: In the given circuit, R1 = 10 Ω, R2 =6 Ω and E = 10 V. Then reading of A2 (A) Reading of A2 is 1 amp (B) Reading of A2 is 2 amp (C) Reading of A2 is 1/4 amp. (D) Reading of A2 is 3 amp Solution: Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be Effective resistance of the circuit Reff = = 20 Ω current through the circuit I = = (1/2) amp Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. ∴ (c) 24: One billion electrons pass from A to B in 1 ms. What is the direction and magnitude of current? (A) 1.6 A (B) 0.8 μA (C) 0.16 μA (D) 1.6 μA Solution : ⇒ i = 0.16 μ A. The current flows from B to A. ∴ (C) 25: An electron gun in a TV set shoots out a beam of electrons. The beam current is 10μA. How many electrons strike the TV screen each second? (A) 2.78 × 1014 (B) 6.3 × 1013 (C) 6.78 × 104 (D) electron will not reach Solution: (B) The number of electrons per second N = I/e = (1.0 × 10–5 C/s)/(1.6 × 10–19 C) = 6.3 × 1013 electrons per second. 26: In above question, how much charge strikes the screen in a minute? (A) –600μC (B) 600μC (C) 1300μC (D) –1300μC Solution: (A) The charge Q striking the screen obeys |Q| = IT = (10μ C/s)(60 s) = 600 μ C. Since the charges are electrons, the actual charge is Q = – 600 μC. 27: A copper bus bar carrying 1200 A has a potential drop of 1.2 mV along 24 cm of its length. What is the resistance per m of the car? (A) 5.2 μΩ (B) 3.2 μΩ (C) 8 μΩ (D) 4.2 μΩ Solution: (D) From Ohm’s law, applied to 24 cm of the bar, V24 = IR24, or (1.2 × 10–3V) = (1200 A) R, and R24 = 1.0 μΩ. By proportion, R100 = (100/24) R24 = 4.2 μΩ 28: A 20–cm–long copper tube has an inner diameter of 0.85 cm and an outer diameter of 1.10 cm. Find its electric resistance when used lengthwise.( ρ = 1.7 × 10–8) (A) 51.2 μΩ (B) 89 μΩ (C) 80 μΩ (D) 42 μΩ Solution: (B) R= ρ(L / A). The cross–sectional area is π[(1.102 – 0.852)]/(4 × 104)] = 3.83 × 10–5 m2; then with L = 0.20 m and ρ = 1.7 × 10–8, then we get R = 89μΩ. 29: At what value of resistance Rx in the circuit in the figure will the total resistance between points A and B is R? (A) R (B) R (C) R (D) R Solution: (A) If Rx = or Rx2 + 2RRx – 2R2 = 0 On solving and rejecting the negative root of the quadratic equation, we have Rx = R 30: What will be the I1 for the following circuit. (A) –0.444 (B) 0 A (C) 0.222 (D) –0.222 Solution: The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their respective 3 Ω) +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C) 31: What will be the I2 for the following circuit. (A) –0.222 (B) –0.212 (C) –0.222 (D) 0.222 Solution: The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their respective 3 Ω) +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C) 32: What will be the I3 for the following circuit. (A) 0.222 (B) –0.444 (C) –0.444 (D) 0.444 Solution: The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their respective 3 Ω) +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C) 33: What will be the I4 for the following circuit. (A) 1 A (B) 0.222 (C) 0 A (D) 2 A Solution: The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their respective 3 Ω) +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C) 34: Three resistances of 4Ω each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between points B and C will be (A) 12 Ω (B) 6 Ω (C) 3 Ω (D) Solution: The circuit can be rearranged as shown in Fig. because it is Wheatstone’s Bridge circuit The resistance between points B and C is given by or R = . Hence the correct choice is (d) 35: What will be the equivalent resistance of the network shown in the figure between the terminals 1 and 2. (A) R (B) R (C) R (D) none Solution: (C) This is the familiar problem, which can be solved using Kirchhoff’s rule. Here, the problem is going to be simplified using Delta–star Transformation. Before going through the solution, let us understand this Delta–Star Transformation. This figure (a) and (b) show the Delta and Star respectively. (a) (b) For obtaining the expression for equivalent we first derive expression for resistance between points of terminals and then equating the expression for resistance of the corresponding terminals of two circuits. Now for Delta Configuration the equivalent resistance between a and c, Ra = Rb = Rc = The given network can be reduced in the form as shown using series and parallel grouping. (c) (d) Using Delta–Star Transformation, the figure (c) can be reduced to figure (d). Now the reduced network is very simple and using series and parallel grouping, the equivalent resistance between terminal 1 and 2 becomes R. CMP I: A cell having a steady emf of 2volt is connected across the potentiometer wire of length 10m. The potentiometer wire is of manganine and having of 11.5 Ω/m. A new resistance of 5 Ω with negligible length is put in series with the potentiometer wire. 36: What is total resistance of wire? (A) 11.5 Ω (B) 115 Ω (C) 1150Ωm (D) 120 Ω Solution: Resistance per unit length of wire = 11.5Ω/m Length of wire = 10 m ∴ Resistance of wire = 11.5 × 10 = 115 ∴ Total resistance = 115 + 5 = 120 Hence correct choice is (D) 37: What is potential gradient when 5Ω resistance is not connected (A) 0.2 V/m (B) 0.4 V/m (C) 0 V/m (D) 1.38 V/m Solution: Fall in potential per meter = = 2/10 = 0.2 v/m ∴ potential gradient = 0.2 V/m Hence correct choice is (A) 38: What is potential gradient after connect 5Ω resistance. (A) 0.2 V/m (B) 0.15 V/m (C) 0.104 V/m (D) 2.0 V/m Solution: Voltage across 115 Ω resistance = Potential gradient = = Hence correct choice is (C) CMP II: An ammeter and a voltmeter are connected in series to a battery with an emf E = 6 volt when a certain resistance is connected in parallel with voltmeter, the reading of latter decreases two times, where as the reading of the ammeter increasing the same number of times. 39: What is ratio of resistance of voltmeter to resistance of ammeter (A) 2 (B) 1/2 (C) 1/3 (D) 3 40: What will be voltmeter reading before the connecting of the resistance. (A) 1 V (B) 2 V (C) 3 V (D) 4 V 41: What will be voltmeter reading after the connecting of the resistance. (A) 1 V (B) 2 V (C) 4 V (D) 3 V Solution: Suppose RA = Resistance of ammeter, Rv = resistance of Voltmeter In the first case current is the circuit ….(1) And voltage across voltmeter V = 6 – Voltage across ammeter V = 6 – iRA ….(2) In the second case reading of ammeter becomes two times i.e. the total resistance become half while the resistance of ammeter remains unchanged. Hence = and voltage across voltmeter V ’ = 6 – i RA …..(3) It is given that … (4) [From (2) and (3)] (Ans. of Question 39) Substituting this value into equation (3) V’ = 6 − 12 × 1/3 ⇒ V’ = 2 (Ans. of Question 40) From Eq, (4) V = 4 (Ans. of Question 41) 42: In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 × 10–11 m with a speed of 2.2 × 106 m/s. Determine its frequency f and the current I in the orbit. (A) 1.06 mA (B) 2.06 mA (C) 0.06 mA (D) 3.06 mA Solution: f = = 6.6 × 1015 rev/s Each time the electron goes around the orbit, it carries a charge e around the loop. The charge passing a point on the loop each second is current = I = ef = (1.6 × 10–19C) (6.6 × 1015s–1) = 1.06 mA. Note that the current flows in the opposite direction to the electron, which is negatively charged. (A) 43: A wire carries a current of 2.0 A. What is the charge that has flowed through its cross–section in 1.0 s. How many electrons does this correspond to? (A) 3.0 C, 1.25 × 1019 (B) 2.0 C, 1.25 × 1019 (C) 4.0 C, 1.25 × 1019 (D) 2.0 C, 5.25 × 1019 Solution: i = ∴ q = it = (2.0 a) (1.0s) = 2.0 C q = ne ∴ n = = 1.25 × 1019 (B) 44: A current of 7.5 A is maintained in a wire for 45s. In this time (a) how much charge and (b) how many electrons flow through the wire? (A) 400.5C, 2.1 × 1021 (B) 300.5C, 2.1 × 1021 (C) 337.5C, 2.1 × 1021 (D) 700.5C, 2.1 × 1021 Solution: (a) q = It = (7.5A) (45 s) = 337.5C (b) The number of electrons N is given by N = = 2.1 × 1021 where e = 1.6 × 10–19 C is the charge of an electron. (C) 45: What will be the current if charge q revolves with the frequency f? (A) f (B) q / f (C) q (D) qf Solution:i = qf (D) 46: Calculate the mean free time between collision in copper at room temperature? and electron average collision per second. (A) 2.4 × 10–14 s (B) 4 × 10–14 s (C) 5.4 × 10–14 s (D) .4 × 10–14 s Solution: n = 8.5 × 1028 m–3 ρ = 1.72 × 10–8 Ω e = 1.60 × 10–19 c and m = 9.11 × 10–31 kg T = = 2.4 × 10–14 s Taking the reciprocal of this time, we find that each electron average about 4 × 1013 collision every second. (A) 47: A coil of wire has a resistance of 25.00Ω at 20°C and a resistance of resistance of 25.17 Ω at 35°C. What is its temperature coefficient of resistance? (A) 4.5 × 10–4°C–1 (B) 5 × 10–4°C–1 (C) 0.5 × 10–4°C–1 (D) 4.0 × 10–4°C–1. Solution: R = Ro[1+ α (T–T0)], or α = ΔR/(R0ΔT), with ΔR = R–R0 = 0.17 Ω and ΩT = T – T0 = 15°C. Then α = (0.17)/(25.00 × 15) = 4.5 × 10–4°C–1. (A) 48: The three resistors in Fig. are R1 = 25 Ω, R2 = 50 Ω, and R3 = 100 Ω. What is the total resistance of the circuit? (A) 50.3 Ω (B) 60.3 Ω (C) 58.3 Ω (D) 80.3 Ω Solution: The sum of R2 and R3 in parallel is or R′ = 33.3 Ω Since R′ is in series with R1, the total resistance of the circuit is R = R′ + R1 = 33.3 Ω + 25 Ω = 58.3 Ω (C) 49: The three resistors in Fig. are R1 = 25 Ω, R2 = 50 Ω, and R3 = 100 Ω. What are the currents l1, l2 and l3 for a 12–V battery? Solution : I = = 0.206 A (A) 0.206 A, 0.137, 0.0685 A (B) 0.506 A, 0.137, 0.0685 A (C) 0.606 A, 0.137, 0.0685 A (D) 0.706 A, 0.137, 0.0685 A The potential V′ across R2 and R3 is V′ = E – R1I = 12 V –(25Ω)(0.206 A) = 6.85 V. Therefore, I2 = = 0.137 A I3 = = 0.0685 A (A) 50: What is total resistance across AB in the following network. (A) 6.4 Ω (B) 2.4Ω (C) 7.4 Ω (D) 5.4Ω Solution: We can’t apply Wheatstone’s bridge condition because ∴ Break delta ΔACD into star Y connection, For that assign 1,2 and 3 like following fig. R1 = = = 0.8 Ω R2 = = = 1.6 Ω R3 = = = 0.8 Ω Connect R1, R2 and R3 like following fig. remove delta circuit from above fig., we get resistance across A and B is R = 0.8 + [(8 + 1.6) || (0.8 + 8)] = = 5.4Ω (D) 51: Find the effective resistance between points A and B of the network shown if Fig. (A) 3 Ω (B) 4 Ω (C) 5 Ω (D) 7 Ω Solution: Resistors AF and FE of 3 Ω each are in series with each other. Therefore, the network AEF is a parallel combination of two 6 Ω resistors. Thus the resistance between points A and E is given by giving RAE = 3 Ω. The network reduces to that shown in Fig. (a). Similarly the resistances between points A (A) 52: Kirchhoff’s current law obeys conservation of (A) charge (B) momentum (C) energy (D) none of these. Solution: (A) 53: The slide wire Wheatstone bridge shown in Fig. is balanced when the uniform slide wire AB is divided as shown. Find the value of the resistance X. (A) 3 Ω (B) 4 Ω (C) 2 Ω (D) 7 Ω Solution: or X = 2Ω (C) 54: A dry cell delivering 2 A has terminal voltage 1.14V. What is the internal resistance of the cell if its open–circuit voltage is 1.59 V? (A) 5.09Ω (B) 6.09Ω (C) 7.09Ω (D) 0.09Ω Solution: The open–circuit voltage is simply the emf of the cell, so V = E – ir with V = 1.41 V, i = 2 A, E = 1.59 V. 1.41 = 1.159 – 2r, and r = 0.09Ω 55. The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times. The shunt used is (A) 88 Ω (B) 14 Ω (C) 6 Ω (D) 16 Ω. Solution : (B) = 30. The shunt S = . 56. The sensitivity of a galvanometer of resistance 8722 Ω is decreased by 90 times. The shunt used is (A) 88 Ω (B) 90 Ω (C) 94 Ω (D) 98 Ω. Solution : (D) Here . 57. A wire l = 8.00 m long, of uniform cross-sectional area A = 8.00 mm2, has a conductance of G = 2.45 Ω–1. What is the resistivity of the material of the wire ? (A) 2.1 × 10–7 S (B) 3.1 × 10–7 S (C) 4.1 × 10–7 S (D) 5.1 × 10–7 S. Solution : (C) 58. A wire 250 cm long and 1 mm2 in cross-section carries a current of 4 A when connected to a 2 V battery. The resistivity of the wire is (A) 0.2 × 10–6 Ω m (B) 2 × 10–7 Ω m (C) 5 × 10–6 Ω m (D) 4 × 10–6 Ω m Solution : (A) R = 59. An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5 Ω. This single wire of cable is replaced by 6 different well-insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to? (A) 7.5 Ω (B) 45 Ω (C) 90 Ω (D) 270 Ω Solution : (A) 60. When cells are arranged in parallel (A) the current capacity decreases (B) the current capacity increases (C) the e.m.f. increases (D) the e.m.f. decreases. Solution : (B) When cells are connected in parallel, the current capacity increases.