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https://docs.google.com/document/d/1R77tXvE5CvMxhRFpWrg095GSL4YS1sn4/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION According to Faraday’s Law, whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. The magnitude of the induced emf is equal to the rate at which flux changes with time. Magnitude of the induced emf = If φ represents the flux through a single turn, and the loop has N turns, then Induced emf = N This induced emf creates an induced current in the circuit whose magnitude is given as i = = LENZ’S LAW According to Lenz’s law, induced emf in a circuit opposes the cause due to which it was induced. Consider the following examples. (a) Suppose that the north-pole of a bar magnet is moved towards a conducting wire loop as shown in the figure. Due to a change in the magnetic flux associated with the loop, a current is induced. Due to induced current, a magnetic field is induced and this magnetic field opposes the motion of bar magnet. The direction of the induced current can be deduced by the following argument: the north pole is moving towards the loop; therefore to oppose the motion of the bar magnet only a north pole will be induced on that face of the loop which faces the magnet. (b) A rectangular loop ABCD is being pulled out of the magnetic field which is directed into the plane of the paper. Perpendicular to the plane of the paper. As the loop is dragged out of the field, the flux associated with the loop decreases. The induced current flows in the loop in a sense so as to oppose the decrease in this flux. For this to happen the magnetic field due to the induced current in the loop must be directed into the plane of the paper. Thus the current in the loop must flow be clockwise. Illustration 1: Magnetic field is increasing into the page with time when a conducting loop of definite radius is placed on the plane of the paper. The find the direction of current in the loop. Solution: As the flux is increasing inside then the current in the loop will be such that it will be opposing the increase in magnetic field, i.e., the induced current in the loop will create such a magnetic field which is directed out ward. Thus the direction of current will be anticlockwise. Illustration 2: A square loop ACDE of area 20cm2 and resistance 5Ω is rotated in a magnetic field = 2T through 180° (a) in 0.01 s and (b) in 0.02 s Find the magnitude of e, i and Δq in both the cases. Solution: Let us take the area vector perpendicular to plane of loop inwards. So initially, ↑↑ and when it is rotated by 180°, ↑↓ Hence, initial flux passing through the loop, φi = BS cos 0° = (2) (20 × 10-4) (1) = 4.0 × 10-3 Wb Flux passing through the loop when it is rotated by 180°, φf = BS cos 180° = (2) (20 × 10-4) (-1) = - 4.0 × 10-3 Wb Therefore, change in flux, ΔφB = φf - φi = - 8.0 × 10-3 Wb (a) Given Δt = 0.01 s, R=5 Ω and Δq = iΔt = 0.16 × 0.01 = 1.6 × 10-3 C (b) Δt = 0.02 s and Δq = iΔt = (0.08) (0.02) = 1.6 × 10-3 C Note: Time interval Δt in part (b) is two limes the time interval in part (a), so e and i are half while Δq is same. Illustration 3: A conducting circular loop with variable radius r is placed in a uniform magnetic field B=0.020T with plane perpendicular to the field. While the radius of the loop is contracting at a constant rate of 1.0 mm/s; find the induced emf in the loop when the radius is 2 mm. Solution: Radius (variable) = r B = 0.02 T Induced emf E = = r = 2 ×10-3 mt E = 0.02 × π× 2 × 2 × 10-3 × 1 × 10-3 = 6 × 10-8 × π = 18.85 ×10-8 V. Illustration 4: A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity ω about Y-axis. The loop lies in a uniform magnetic field B in the direction of X-axis. Assuming that at t = 0, the plane of the loop is normal to the lines of force, find an expression for the peak value of the emf and current Induced in the loop. What is the magnitude of torque required on the loop to keep it moving with constant ω? Solution: As φ is maximum at t = 0, φ (t) = BA cos ωt Magnitude of induced emf = N Magnitude of induced current = (a) ⇒ peak value of emf = BAωN peak value of induced current = BAωN/R (b) Power input = heat dissipation per sec ⇒ ⇒ MOTIONAL EMF Induced emf due to motion of a conductor in magnetic field Suppose a conductor is moving with velocity in a uniform magnetic field . We take a small element PQ = d of the conductor, then induced e.m.f. in the element. or, e = (vB sinθ). d cosα Where θ is the angle between and and α is the angle between element and direction of force Illustration 5: A metal rod 1.5 m long rotates about its one end in a vertical plane at right angles to the magnetic meridian. If the frequency of rotation is 20 rev/s. Find the emf induced between the ends of rod. BH = 0.32 × 10-4T: Horizontal component of earth’s magnetic field. Solution: Length of rod = 1.5 mt. Frequency of rotation = 20 rev/sec. BH = 0.32 × 10-4 T Angular velocity of rod = 20 × 2π rad/sec. emf induced = = 0.32 × 10-4 × 20 × π× (1.5)2 = 4.5 × 10-3 V INDUCTANCE SELF INDUCTION We have already discussed capacitors – devices that store energy using electric fields. Like a capacitor, an inductor is also quite a commonly used element in electric circuits. It stores magnetic energy. As we know that when current flows through a conductor a magnetic field is set-up in surrounding of it, and hence it is associated with magnetic flux. If magnetic flux associated with a coil is φ and current flowing through it is I, then its inductance is given by the expression. The quantity 'L' is called self-inductance of the coil. It does not depend on the current, but it depends on the permeability of the core and the dimensions of the coil. S.I. unit of inductance is Henry. Consider the circuit, in which a solenoid is connected across a cell through a resistor. When the switch is open, the current in the circuit is zero. When the switch is closed, a current flow in it. Since current in the circuit increases from zero to a certain value, magnetic field associated with it changes that causes induction of an emf across the solenoid. Induction of an emf due to variation in current flowing through the coil itself is known as self induction. Since φB = LI, and ε = . Inductance of an ideal solenoid: Let a current I flow through a solenoid. The magnetic field due to the current flowing within the solenoid is, B = μ0nI, where n is the number of turns per unit length. If area of cross section of the solenoid is A then flux associated with length  is equal to φ = nBA. (Assuming that the solenoid is ideal and long) where  is the length of the solenoid. Now B = μnI ⇒ = μ0 n2A Self Inductance of a Coil Consider a coil of N turns and area of crossection A carrying a current i. The length of the coil is √A). Comparing with φ = L i, we get: L = R - L CIRCUIT GROWTH OF CURRENT When switch is on at t = 0. Then current starts from 0 to I. And self inductor opposes the grow of current ⇒ ⇒ In ⇒ I = potential difference across inductor = I = At t = 0, I = 0 At t → ∞ ⇒ I → V = At t = 0 ⇒ potential difference across inductor = ε At t → ∞ ⇒ potential difference across inductor = 0. So inductor will behave like plane wire. Consider graph betn I and time. Consider graph betn potential difference across Inductor and time Illustration 6: A coil of inductance 1.0 H and resistance 100 Ω is connected to a battery of emf 12 V. Find the energy stored in the magnetic field of the coil 10 ms after the circuit is switched on. Solution: L = 1.0 H R = 100 Ω E = 12 V i(t) = Energy stored in the magnetic field is 1/2 L i2. = = Illustration 7: A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. is induced in the coil. Solution: For induced emf to develop in a coil the magnetic flux through it must change. But in this case the number of magnetic lines of force through the coil is not changing. Therefore the statement is false. INDUCTANCE OF COMMON ELEMENTS 1. Wire L = 2. Hollow cylinder L = >> a 3. Parallel wires L = >> d, d >> a 4. Coaxial conductor L = 5. Circular loop L = = –2 ρo,ρo >> d 6. Solenoid L = >> a 7. Torus (of circular cross section) L = 0N2[ρo – ] 8. Sheet L = o 2 MUTUAL INDUCTION Consider two coils C1 and C2 placed as shown. By varying current i1 in coil C1, we change the flux not only through C1 but also through coil C2. The change in flux φ2 through C2 (due to change in current i1) induces an emf in the coil C2. This emf is known as mutually induced emf and the process is known as mutual induction. φ2 ∝ i1 ⇒ φ2 = M i1 Where M is called as mutual inductance of the pair of coils. The coil C1 in which i varies is often called primary coil and the coil C2 in which the emf is induced is called secondary coil. induced emf in coil C2 = E2 The mutual inductance is maximum when the coils arc wound up on the same axis. It is minimum when the axes of coils are normal to each other. Note down the following points regarding the mutual inductance: 1. The SI unit of mutual inductance is henry (1H). 2. M depends upon closeness of the two circuits, their orientations and sizes and the number of turns etc. 3. Reciprocity theorem: M21 = M12 = M e2 = - M(di1/dt) and e1 = - M(di2/dt) 4. A good approach for calculating the mutual inductance of two circuits consists of the following steps: (a) Assume any one of the circuits as primary (first) and the other as secondary (second). (b) Suppose a current i1 flows through the primary circuit. (c) Determine the magnetic field produced by the current i1. (d) Obtain the magnetic flux . (e) With the flux known, the mutual inductance can be found from, Illustration 8: A long solenoid of length 1 m, cross sectional area 10 cm2, having 1000 turns has wound about its centre a small coil of 20 turns. Compute the mutual inductance of the two circuits. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s? Solution: Let N1 = number of turns in solenoid; N2 = number of turns in coil A1 and A2 be their respective areas of crossection. (A1 = A2 in this problem) Flux φ2 through coil crated by current i1 in solenoid is φ2 = N2(B1A2) ⇒ Comparing with φ2 = M i1; we get Mutual inductance = M = Magnitude of induced emf = E2 = M Illustration 9: The coefficient of mutual induction between the primary and secondary of a transformer is 5 H calculate the induced emf in the secondary when 3 Amp current in the primary is cut-off in 2.5 × 10-4 sec. Solution: M = coefficient of mutual induction = 5 H. 3 Amp current is a cut off in 2.5 × 10-4 sec. E = = 6 ×104 V L–C CIRCUIT L.C. OSCILLATIONS A capacitor is charged to a potential difference of Vo by connecting it across a battery and then is allowed to discharge through an inductor of inductance L. Initial charge on the plates of the capacitor qo = CVo At any instant, let the charge flown through the circuit be q and the current in the circuit be i. Applying Kirchhoff’s law − = 0 Differentiating w.r.t. time we get = 0 = = −ω2i, f = ∴ The charge q on the plates of the capacitor and current I in the circuit vary sinusoidally as q = q0 sin (ωt + φ) and I = q0 ω cos (ωt + φ). where φ is the initial phase and it depends on initial conditions of the circuit and ω = . The total energy of the system remains conserved = constant = Illustration 10: A–2–F capacitor is initially charged to 20 V and then shorted across a 6–H inductor. What are the frequency of oscillation and the maximum value of the current? Solution: The frequency of oscillation is independent of the initial charge and depends only on the values of the capacitance and inductance. The frequency is f = = = 4.59 × 104 Hz According to equation the maximum value of the current is related to the maximum value of the charge by Im = ωQ0 = The initial charge on the capacitor is Q0 = CV0 = (2 F)(20 V) = 40 C Thus Im = = 11.5 A MASS–SPRING SYSTEM VS INDUCTOR–CAPACITOR CIRCUIT A comparison of oscillations of a mass spring system and an L - C circuit. S. No. Mass Spring System Inductor-Capacitor Circuit 1. Displacement (x) Charge (q) 2. Velocity (v) Current (i) 3. Acceleration (a) Rate of change of current 4. 5. x = A sin (ωt ± φ) or x = A cos (ωt ± φ) q = q0 sin (ωt ± φ) or q = q0 cos (ωt ± φ) 6. 7. Rate of change of current 8. Kinetic energy = mv2 Magnetic energy = Li 9. Potential energy = kx Potential energy = 10. mv2 + kx2 = constant = kA2 = Li2 + = constant = = 11. imax = q0ω 12. 13. C 14. m L ALTERNATING CURRENTS AND CIRCUITS The basic principle of the ac generator is a direct consequence of Faraday’s law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency ω a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is, V = V0 sin ωt (i) The usual circuit diagram symbol for an ac source is shown in Figure. In Equation (i) V0 is the maximum output voltage of the ac generator, or the voltage amplitude and ω is the angular frequency, equal to 2π times the frequency f. ω = 2πf The frequency of ac in India is 50 Hz, i.e., f = 50 Hz So, ω = 2πf ≈ 314 rad/s The time of one cycle is known as time period T, the number of cycles per second the frequency f. A sinusoidal current might be described as, i = i0 sin ωt If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is, Thus, Similarly, the average value of the voltage (or emf) for one cycle is zero. Since, these averages for the whole cycle are zero, the dc instrument will indicate zero deflection. In ac, the average value of current is defined as its average taken over half the cycle. Hence, This is sometimes simply written as, iav. Hence, Similarly, A dc meter can be used in an ac circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, i.e., time the maximum current i0 . A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i2 and finally take the square root of that average. This procedure defines the root-mean-square current denoted as irms. Even when i is negative, i2 is always positive so irms is never zero (unless i is zero at every instant). Hence, Thus, Similarly, we get The square root of the mean square value is called the virtual value and is the value give by ac instruments. Thus, when we speak of our house hold power supply as 220 volts ac, this means that the rms voltage is 220 volts and its voltage amplitude is, V0 = Vrms = 311 volt Form Factor The ratio, is known as or factor, Note: 1. The average value of sin ωt, cos ωt, sin2ωt, cos2ωt, etc. is zero because it is positive half of the time and negative rest half of the time. Thus, ∴ If i = i0 sin ωt then 2. The average value of sin2 ωt and cos2 ωt is or ∴ If then 3. Like SHM, general expressions of current/voltage in an sinusoidal ac are, i = i0 sin (ωt ± φ), V = V0 sin (ωt ± φ) or i = i0 cos (ωt ± φ), and V = V0 cos (ωt ± φ) Illustration 11: If the current in an ac circuit is represented by the equation, i = 5 sin (300t - π/4) Here, t is in second and i in ampere. Calculate, (a) peak and rms value of current. (b) frequency of ac (c) average current Solution: (a) An in case of ac, i = i0 sin (ωt ± φ) ∴ The peak value i0 = 5A and (b) Angular frequency ω = 300 rad/s ∴ (c) Capacitor in an AC Circuit If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor, q = CVC = CV0 sin ωt and the instantaneous current i passing through it, is given by: or i = i0 sin (ωt + π/2) Here, This relation shows that the quantity is the effective ac resistance or the capacitive reactance of the capacitor and is represented as XC. It has unit as ohm. Thus, It is clear that the current leads the voltage by 90° or the potential drop across the capacitor lags the current passing it by 90°. Figure shows V and i as functions of time t. Inductor in an AC Circuit Consider a pure inductor of self inductance L and zero resistance connected to an alternation source. Again we assume that an instantaneous current i = i0 sin ωt flows through the inductor. Although there is no resistance, there is a potential difference VL between the inductor terminals a and b because the current varies with time, giving rise to self induced emf. VL = Vab = - (induced emg) = Or VL = Or VL= V0 sin (i) Here V0 = i0(ωL) (ii) Or i0 = ∴ (iii) Equation (iii) shows that effective ac resistance, i.e., inductive reactance of inductor is, XL = ωL And the maximum current, The unit of XL is also ohm. From Equations. (i) and (iii) we see that the voltage across the inductor leads the current passing through it by 90°. Figure shows VL and i as functions of time. Illustration 12: A 100Ω resistance is connected in series with a 4H inductor. The voltage across the resistor is, VR = (2.0V) sin (103 rad/s)t: (a) Find the expression of circuit current (b) Find the inductive reactance (c) Derive an expression for the voltage across the inductor. Solution: (a) (2.0 × 10-2 A) sin (103 rad/s)t (b) XL = ωL = (103 rad/s) (4H) = 4.0 × 103 ohm (c) The amplitude of voltage across inductor, V0 = i0XL = (2.0 × 10-2 A) (4.0 × 103 ohm) = 80 volt In an ac voltage across the inductor leads the current by 90° or π/2 rad. Hence, VL = V0 sin (ωt + π/2)= (80 volt) sin Note: That the amplitude of voltage across the resistor (=2.0 volt) is not same as the amplitude of the voltage across the inductor (=80 volt), even though the amplitude of the current through both devices is the same. SERIES L–R CIRCUIT As we know potential difference across a resistance in ac is in phase with current and it leads in phase by 90° with current across the inductor. SERIES C–R CIRCUIT Potential difference across a capacitor in ac lags in phase by 90° with the current in the circuit. Suppose in phasor diagram current is taken along positive x-direction. Then VR is also along positive x-direction but VC is along negative y-direction. So we can write. V = VR – jVC = iR – j(iXC) =iR – j =iZ Here, impedence is, Z = R - j The medulus of impedance is, and the potential difference lags the current by an angle, = or SERIES L–C–R CIRCUIT Potential difference across an inductor leads the current by 90° in phase while that across a capacitor, it lags in phase by 90°. Suppose in a phasor diagram current is taken along positive x-direction. Then VR is along positive x-direction, Then VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction. So, we can write, V = VR + jVL – jVC = iR + j(iXL) – j(iXC) = iR + j [ i (XL – XC)] = iZ Here impedence is, Z = R + j(XL – XC) = R + j The modulus of impedence is, And the potential difference leads the current by an angle, or Illustration 13: An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1Ω and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Also find the virtual current in the circuit. Solution: In case of an ac, the voltage leads the current in phase by an angle, Here, XL = ωL = (2πfL) = (2π) (50) (0.01) = πΩ and R = 1Ω ∴ φ = tan-1 (π) ≈ 72.3° Further, irms = Substituting the values we have, irms = 60.67 amp. Illustration 14: Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values 1000Ω, 1μF nd 2. henry respectively. Given emf as, V = sin 1000 t volt Solution: The rms value of voltage across the source, ω = 1000 rad/s ∴ irms = = 0.0707 amp The current will be same every where in the circuit, therefore, P.D across resistor VR = irmsR = 0.0707 × 1000 = 70.7 volt P.D across inductor VL = irmsXL = 0.0707 × 1000 × 2 = 141.4 volt and P.D across capacitor VC = irmsXC = 0.0707 × = 70.7 volt Note: (i) The rms voltages do not add directly as, VR + VL + VC = 282.8 volt Which is not the source voltage 100 volts. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra. CIRCUIT ELEMENTS WITH AC Circuit elements Amplitude relation Circuit quantity Phase of V Resistor V0 = i0R R In phase with i Capacitor V0 = i0XC Lags i by 90° Inductor V0 = i0XL XL = ωL Leads i by 90° ASSIGNMENT 1. In a series resonant circuit the ac voltages across resistance R, inductance L and capacitance C are 5V, 10V and 10V respectively. The ac voltage applied to the circuit will be (A) 25 V (B) 20 V (C) 10 V (D) 5 V. Solution: (D) V = [ + (VL – VC)2]1/2. 2. There is a current of 20 milliampere through an ideal choke coil of 1.5 H when connected to 220 volt, 50 hertz ac supply. The power consumed is: (A) Zero (B) 220 × 20 × 10–3 watt (C) 220 × 50 watt (D) 220 × 20 × 1.5 watt. Solution: (A) Ideal choke is a wattles device. Because the phase angle between the current and efm is . This gives power factor cosφ = cos = 0. 3. The ratio of induced emf in a coil of 50 turns and area A oscillating at frequency 50 Hz to that in a coil of 100 turns and same area oscillating at frequency 100 Hz is : (A) 0.25 (B) 0.50 (C) 0.75 (D) 1.00 Solution: (A) E ∝ nABω. Therefore . 4. In the above questions what is the potential drop across the capacitor? (A) 110 V (B) 110 V (C) 220 V (D) 220 V. Solution: (A) AT resonance, the potential drop across the capacitance is equal and opposite to that inductance. 5. The rms current in an a.c. circuit is 2 A. If the wattles current be A, what is the power factor? (A) (B) (C) (D) . Solution: (C) Iwattless =Irms sinφ Hence sin φ = ⇒ φ = 60 Therefore cos φ = . 6. A circular wire loop of radius r can withstand a radial force T before breaking. A particle of mass m and charge q(q > 0) is sliding over the wire. A magnetic field B is applied normal to the plane of the wire. The maximum speed Vmax the particle can have before the loop breaks is (A) Zero (B) (C) (D) Solution: (D) The following forces act on the particle. ∙ Force T acting radially inwards. ∙ Centrifugal force acting radially outwards. ∙ Magnetic force qvB acting radially inwards. ⇒ ⇒ ⇒ ⇒ v = ⇒ 7. A rectangular loop of sides a and b, has a resistance R and lies at a distance c from an infinite straight wire carrying current I0. The current decreases to zero in time τ , 0 < t < τ The charge flowing through the rectangular loop is (A) μ0I0τ (B) (C) (D) Solution: (C) Consider an infinitesimal element of length b. thickness dx at a distance x from the wire. Since dφ = BdA ⇒ ⇒ ⇒ ⇒ Since ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ 8. A network of inductances, each of value 1H, is shown in figure. The equivalent inductance of the circuit between points A and B is (A) 6.218 H (B) 0.268 H (C) 8.162 H (D) 2.618 H Solution: (D) 1 and 2 in series in parallel with 3. The result of this is in series with 4, which is in parallel with 5, again in series with 6, which again is in parallel with 7, in series with 8 and in parallel with 9. Finally result of above steps, 10 & 11 are in series to get 2.618 H. 9. A uniformly wound solenoid coil of self inductance 1.8 × 10–4 henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-voll battery of negligible resistance. The time constant for the current in the circuit and the steady state current through the battery is (A) 0.3 × 10-4S, 8A (B) 1.3 × 10-4S, 8A (C) 0.3 × 10-4S, 5A (D) none of these Solution: (A) (i) L, R The coil is broken into two identical coils. , Time constant = Steady current I = 10. If ε0 and μ0 are, respectively, the electric permittivity and magnetic permeability of free space, ε and μ the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is (A) 2 (B) (C) (D) none of these Solution: (B) We know that the velocity of light in vaccum And the velocity of light in a medium Also the refractive index 11. The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A, and is decreasing at a rate of 103 A/s then VB – VA = (A) 15V (B) 20V (C) 25V (D) 30V Solution: (A) ⇒ VB – VA = 15 – e – I Here, I = 5A, VB – VA = 15V 12. What is the nature of RCL series circuit for frequency less than the resonant frequency? (A) Resistive (B) Inductive (C) Capacitive (D) None of the above Solution: (C) For ω < ω0, XL < XC. So, the circuit is capacitive. 13. If the power factor changes from to then what is the increase in impedance in A.C.? (A) 20 % (B) 50 % (C) 25 % (D) 100 % Solution: (D) cos ∝ Z2 = 2Z1 Percentage change = × 100 = 100 %. 14. What is the impedance of a purely antiresonant circuit at resonance? (A) Zero (B) R (C) 2 R (D) . Solution: (D) For antiresonant circuit Z = . For purely antiresonant circuit R = 0. Hence Z = . 15. A small dc motors operating at 200 V draws a current of 3 A at its full speed of 2500 r.p.m. If the resistance of the armature is 7Ω, the back e.m.f. of the motor is (A) 149 V (B) 159 V (C) 169 V (D) 179 V Solution : (D) E = V-I R = 200-3x7 = 179 V. 16. A motor (A) converts electrical energy into mechanical energy (B) converts mechanical energy into electrical energy (C) converts electrical energy into magnetic energy (D) produces mechanical energy. Solution : (A) 17. For maximum output power in D.C. motor, the induced back emf (E) should be (A) applied voltage (B) double applied voltage (C) half of applied voltage (D) one third of applied voltage Solution (C) For maximum output, E = V / 2 18. In ac motor, a capacitor is used to (A) reduce ripple (B) reduce A.C. (C) introduce phase difference of (D) increase A.C. Solution : (C) The capacitor introduces a phase difference of π/2 which is necessary for motors. 19. A generator develops an e.m.f. of 130 V and has a terminal potential difference of 125 V, when the armature current is 25 A. The resistance of the armature is (A) 0.5 Ω (B) 0.2 Ω (C) 1.5 Ω (D) 2.4 Ω Solution : (B) 20. A coil is wound of a frame of rectangular cross-section. If the linear dimensions of the frame are doubled and the number of turns per unit length of the coil remains the same, then the self inductance increases by a factor of (A) 6 (B) 12 (C) 8 (D) 16. Solution : (C) The number of turns has become twice and area has gone up four times. Therefore self inductance is 4x2 = 8 times. XL = ωL = 2π vL with iron core XL increases. Since , I decreases. 21. A Circuit contains two inductors of self inductance L1 and L2 in series. If M is the mutual inductance, then the effective inductance of the circuit shown will be (A) L1+L2 (B) L1+L2 -2M (C) L1+L2 +M (D) L1+L2 +2M Solution : (D) when inductances are connected like this in series, then L = L1+L2+2M. 22. A circuit contains two inductors of self inductance L1 and L2 in series as shown in fig. If M is the mutual inductance then the effective inductance of the circuit is (A) L1+L2 (B) L1+L2 -2M (C) L1+L2 +M (D) L1+L2 +2M Solution : (B) in this type of combination, L = L1+L2-2M 23. Two coils of self inductance L1 and L2 are in parallels as shown in fig. If their Mutual inductance is M, then the effective inductance will be given by (A) (B) (C) (D) Solution : (D) When two inductances are in parallels then L = 24. Air cored chokes are used for reducing high frequency ac because (A) air is free (B) air core has high permeability (C) with air core self inductance increases (D) with air core self inductance is not much. Solution : (D) XL = ω L = 2π vL, since v is large therefore L should not be large. Hence air cored chokes are used for high frequency ac. 25. In an ac circuit V and I are given by V = 150 sin (150 t) V and I = 150 sin A The power dissipated in the circuit is (A) 5625 W (B) 4825 W (C) 7450 W (D) 3425 W Solution : (A) E0 = 150 V and I0 = 150 x 10–3 mA and φ = 600 26. The induced e.m.f. in a coil rotating in a uniform magnetic field depends upon (A) only on total number of turns in the coil (B) only on magnetic field (C) area of coil and speed of rotation (D) all of the above Solution : (D) E = n AB ω, so it depends on all these. 27. A resistance and an inductor are connected in series to a 220 V A.C. supply. When measured with an A.C. voltmeter, the P.D. across resistor is 140 V. The P.D. across the terminals of the inductor will be nearly (A) 150 V (B) 160 V (C) 170 V (D) 180 V solution (C) 220 = or 48400 = 19600 + VL2 VL = = 170 V 28. If q0 is the maximum charge then while charging, the charge on a capacitor at time t is given by (A) q0 t (B) (C) (D) Solution : (D) q = where RC is the time constant. 29. In the case of charging of a capacitor, at t = RC, the charge on the capacitor (q0 is the maximum charge) is (A) 0.432 q0 (B) 0.532 q0 (C) 0.632 q0 (D) 0.732 q0 Solution : (C) When t = RC, q = q0 (1-e–1) = 0.6321 q0 30. In the case of charging or discharging of a capacitor RC has the units of (A) Ohm (B) A-F (C) Ohm - m (D) sec. solution : (D) The quantity RC has units of For test 1. The capacitor gets almost fully charged after time t equal to (A) RC (B) 3 RC (C) 4 RC (D) 5 RC Solution : (D) After t = 5 RC, the capacitor is about 99.3% charged. 2. The time constant for the given RC circuit where R = 1 K Ω and C = 103 pF will be (A) 2 m second (B) 1 μ second (C) 1.5 m second (D) 0.5 μ second. Solution : (B) t = RC = 103 x 103 x 10–12 = 10–6 s = 1μs 3. A capacitor of 4μF is connected to a 15 V supply through 1 mega ohm resistance. The time taken by the capacitor to charge upto 63.2 % of its final charge will be (A) 2 s (B) 3 s (C) 4 s (D) 5 s. Solution : (A) 63.2 % = RC = 106 x 4 x 10–6 = 4 s 4. If ∝ is a real number, cos ∝ + j ∝ is a complex number where j is equal to (A) 1 (B) 1 (C) –1 (D) Solution : (D) i = 5. The Exponential function is equal to (A) cos θ (B) cos θ + sin θ (C) cos θ - j sin θ (D) cos θ + j sin θ Solution : (D) = cos θ + j sin θ 6. If LCR are connected is series, then the total effective resistance called impedance in terms of complex number is giver by (A) (B) (C) (D) Solution : (D) Z = 7. The magnetic flux through a coil perpendicular to its plane and directed into paper is varying according to relation = 5t2 + 10t + 5 milli weber. The e.m.f. induced in the loop after 5 sec 15 (A) 0.03 volt (B) 0.06 volt (C) 0.08 volt (D) 0.02 volt Solution : (B) = (5t2 + 10t + 5) x 10–3 Wb as e = (in magnitude) (5t2 + 10t + 5) x 10–3 Wb sec–1 = (10t + 10) 10–3 volt e = (10 x 5 + 10) x 10-3 = 0.06 volt 8. A train is moving with a speed of 30 m s–1 NS on the rails separated by 2 m. If the vertical component of earth's field is 8 x 10–5 T, the e.m.f. is (A) 0.0048 V (B) 0.048 V (C) 0.48 V (D) 4.8 V Solution : (A) e = Blv = 30 x 2 x 8 x 10-5 = 0.0048 V 9. An aircraft with a wingspan of 40 m flies with a speed of 1080 km h–1 in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth's magnetic field is 1.75 x 10–5 T. Then the e.m.f. that develops between the tips of the wings is (A) 0.5 V (B) 0.34 V (C) 0.21 V (D) 2.1 V Solution : (C) L = 40 mv = 1080 km h-1 = 300 m sec-1 B = 1.75 x 10-5 T e = Blv = 1.75 x 10-5 x 40 x300 = 0.21 V 10. A Uniform flux density of 0.1 Wb m2 extends over a plane circuit of area 1 m2, and is normal to it. How quickly must the field be reduced to zero if an emf of 100 V is to be produced? (A) 10–2 Sec (B) 10–3 sec (C) (D) 1 micro sec. Solution : (B) = BA = 0.1 x 1 = 0.1 Wb; = 0 e = - 100 = , t = 10-3 s 11. Current in a 10 milli henry coil increases uniformly form zero to one to one ampere in 0.01 second, then the value of induced e.m.f. is (A) + 1 V (B) 2 V (C) –1 V (D) –2 V Solution : (C) L = 10 x 10-3 = 10-2 henry change in current = 1-0 = 1 amp. dt = 0.01 sec e = - = -10-2 x = -1 volt 12. A step up transformer is used on 120 V line to provide a P.D. of 2400 V. If the number of turns in primary is 75, then the number of turns in the secondary shall be (A) 25 (B) 150 (C) 1500 (D) 500 Solution : (C) Here EP = 120 V, ES = 24000 V, np = 75 and ns = ? As = 13. The current drawn from the primary of a transformer which steps down 220 V to 22 V at 0.1 amp. shall be (A) 1 amp. (B) 0.1 amp (C) 0.2 amp. (D) 0.01 amp Solution : (D) Here np = 220 V, ns = 22 V, Is = 0.1 amp. and Ip = ? 14. What is the r.m.s. value of current ? I = (A) (B) 1 A (C) (D) 2 A. Solution : (B) Peak value of current, I0 = A 15. A hot wire instrument reads 15 A. The peak value of the current is (A) (B) (B) 15 A (D) Solution : (A) The hot wire instruments measure virtual values only. 16. The exponential function ex is equal to (A) x+x2+x3+.... (B) (C) (D) Solution : (B) ex = 17. An A.C. voltmeter reads 20 V across a resister, 30 V across an inductor and 15 V across a capacitor in LCR series circuit. In this case the supply voltage will be (A) 100 V (B) 60 V (C) 50 V (D) 25 V Solution : (D) e = = 18. An dc voltage, E (in volt) = 200 sin 100 t is connected through an dc ammeter. The reading of the ammeter shall be (A) 10 mA (B) 20 mA (C) 40 mA (D) 80 mA Solution : (B) 19. An LCR series circuit its connected to a source of A.C. at resonance, the applied voltage and the current flowing through the circuit will have a phase difference of (A) (B) (C) (D) 0. Solution : (D) At resonance 20. The Capacitor shown in the Fig. is initially charged to a voltage V and the circuit is then closed. The charge on C is found to oscillate with frequency w. The frequency will be doubled if (A) the voltage V is doubled (B) both L and C are halved (C) both L and C are doubled (D) both L and C are creased by a factor Solution : (B) . 21. In a parallel LCR circuit shown in the Fig. at resonance (A) the source current is maximum (B) the impedance of the circuit is minimum and is equal to R (C) the resonance frequency will be the same as for a series resonance circuit with the same value of L, C and R (D) the voltage across L and C are in phase. Solution : (C) In a parallel resonance circuit, at resonance for a good qual;ity capacitor Rc = 0 and we put RL = R. Then If resistance is neglected then condition of resonance in both series and parallel circuit become the same, namely 22. In a series LCR ac circuit, off resonance, the current is (A) always in phase with the generator voltage. (B) always lags the generator voltage (C) always leads the generator voltage (D) may lead or lag behind the generator voltage. Solution : (D) In a series LCR circuit, off resonance, the power factor is So, Now depending on whether is greater than or less, the lead or lag will occur. 23. Who discovered laws of EMI ? (A) Faraday (B) Einstein (C) P.A.M. Dirac (D) R.P. Feynman. Solution : (A) 24. The Dimensions of flux is (A) MLT-2 A (B) MLT-2A-1 (C) ML-1T-3A-1 (D) ML2T-2A-1 Solution : (D) ( F = Bqv) 25. When a current of 4 A through primary gives rise to a flux of magnitude 1.35 Wb through secondary, what is the coefficient of mutual induction (M)? (A) 0.43 H (B) 0.38 H (B) 0.34 H (D) None of these. Solution : (C) 26. When current changes from 13 A to 7 A in 0.5 second through a coil, the e.m.f. induced is 3 x 10-4 V. What is the coefficient of self induction? (A) 25 x 10-3 H (B) 25 x 10-4 H (C) 25 x 10-5 H (D) 25 x 10-6 H Solution : (D) e = 300 x 10–6 V dt = 0.5 s e = L in magnitude = 25 x 10-6 H 27. For two coils inductively coupled, the mutual inductance is 1.9 H. The current through primary changes by 6 A in 1 ms. The induced e.m.f. is (A) 11.4 mV (B) 11.4 V (B) 11.4 kV (D) Solution : (A) M = 1.9 H, dI = 6 A, dt = 10-3 sec = 11.40 x 10-3 V 28. In the middle of a long solenoid there is a coaxial ring of square cross-section, made of conductivity material of resistivity p. The thickness of the ring is equal to h, its inside and outside radii are equal to 'a' and 'b' respectively. What is the current induced in a radial width (dr), where the magnetic field varies with time as (A) (B) (C) (D) Solution : (D) According to the problem, the e.m.f. induced in an elementary ring of radius r and width dr is . The conductance of this ring is thus, the current induced is dI = . Upon integration we get the total current 29. In the Q.95, what is the total current ? (A) (B) (C) (D) Solution : (A) According to the problem, the e.m.f. induced in an elementary ring of radius r and width dr is . The conductance of this ring is thus, the current induced is dI = . Upon integration we get the total current 30. Suppose p is the mass density, po is the resistivity, l0 is the length and S is the cross-section of a wire. The mass of the wire is m. What is the active resistance of the wire ? (A) (B) (C) (D) Solution: (A) The time constant τ is the ratio L/R, where L is the inductance of the solenoid and R is the active resistance. The latter is given by R = ρ0l0/S where ρ0 is the resistivity, l0 is the length and S is the cross-section of the wire. Also, if ρ is the mass density of the material of the wire, mass of the wire m = ρl0S. Upon eliminating S one can write R = ρ0 l0 / S The time constant (τ) is therefore given by The length of the wire l0, in terms of the length of the solenoid l, is given by Thus, the time constant τ is given by