Current Electricity-03-SUBJECTIVE SOLVED

SOLVED SUBJECTIVE 1. What’s the effective resistance of following circuits? R (a) (b) R C D Solution:(a) It is a Wheatstone bridge that is balanced. Hence the central resistance labeled ‘C’ can be pulled out.  Req  R (b) The R11 is in parallel with a balanced Wheatstone bridge.  R  R.R  R . eq R  R 2 2. An ohm-meter is used for measuring the resistance of a circuit between two terminalsof given circuit. (all resistances are in Ohm) What would be the reading of such an instrument used for the circuit between points (a) AB (b) AC (c) BC Solution: (a)   = 25  (b)  1 1 1 R = 40 + 60  R = 24  (C) 1 1 1 1 6  3 1 R = 15 + 30 + 90 = 90  R = 9  . A 3. B (a) Find i and i1 in the shown network, (b) Poential difference across A - B Solution: (a) Req =6  24 i = 6 = 4A i 4 i1 = 16 = 16 i = 0.25A (b) VAB = 4 × 4 4 = 4 × 4 = 4 volt. 4. Using Kirchoff’s current law and ohm’s law. Find the magnitude and polarity of voltage V. Direction of current is as shown. at point A Solution: i1 + i2 + i3 + 8 – 30 = 0 V i1 = 2 V i2 = 6 V i3 = 4 V V 2 + 6 V + 4 + 8 = 30 V  1  1  1  = 22  2 6 4  V623 = 22 decreases from A to B.  12  V = 24 volt i1 = 12 A i2 = 4A, i3 = 6A, 5.   Find VCE and VAG 20 Solution: i1 = 20 40 i2 = 20 = 1A = 2A VCE  VC  VE  VC  VB  (VB  VH )  (VH  VE )  10  10  5  5V VAG = VA  VG  (VA  VB )  (VB  VH )  (VH  VG )  6 1  10  7  2  30V . 6. Solve for the power delivered to the 10  resistor in the circuit shown.  1 Solution: 10 15 (Using super-position pirinciple) 15 2 6 2 6 I = 20 1 3 = 23 , I1 = 3 × 23 = 0.17  1 V I = 10 11 60  5 15 = 660 66 230 23 I1 = 1 66 11 × 23 6 = 23 = 0.26 Net current in 10  = 0.17 + 0.26 = 0.43 p = i2R = (0.43)2 × 10 = 1.85 W 7. Find the effective resistance of netwrok between A & B Solution: A & B are symmetric points Now current in A to C is same as from D to B and similarlly A to E to B has to be same  E is no true point 8 r . 2 r 3 = 8 r  2 r 3 2r 16r 3 14r 3 8. Find potential at A, B, C, D, E (i) if A is grounded (ii) If D is grounded. Solution: Potential at A = 0 at B = 34 at C = 34 – 4 = 30 at D = 30 – 10 = 20 at E = 20 – 8 = 12 Potential at D = 0 at E = – 8 volt at A = – 20 volt at B = 14 volt at C = 14 – 4 = 10 volt. 9. An ammeter of resistance 0.80  can measure currents upto 1.0A. (a) What must be the shunt resistance to enable the ammeter to measure current upto 5.0A? (b) What is the combined resistance of the ammeter and the shunt? Solution: Maximum current which can be passed through the galvanometer Ig = 1.0A. (a) Total current in the circuit I = 5.0A Shunt resistance S = ? We know I  I  S  G  5   S  0.8   1  0.8 g  S  1 S  S     or S  .8  0.2 4 (b) The combined resistance will be 1  1  1 1  1 4 R 0.8 0.2 or R  0.16 . R 0.8 10. A series battery of 6 lead accumulators each of emf 2.0 V and internal resistance 0.25  is charged by a 230 V dc mains. To limit the charging current, a series reistance of 53  is used in the charging circuit. What is (a) the power supplied by the mains, and (b) the power dissipated as heat? Account for the difference in the two answers. Solution: The emf of the battery  6  2.0V  12V Internal resistance of the battery  6  0.25  1.5 Charing current  (230 12)V (53  1.5) = 4.0 A Power supplied by the mains  230V  4.0A  920W Power dissipated as heat  42  (53 1.5)W  872W The difference (920 – 872) W =48 W, is the power stored in the accumulator in the form of chemical energy of its constents. 11. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 Ω (fig.). Determine the equivalent resistance of the network and the current along each edge of the cube. D I/2 A I C I I/2 B I/2 I I D A 3I I/2 I I C I/2 B I/2 E 10V Solution: The network is not reducible to a simple series and parallel combinations of resistances. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network. The paths AA’, AD and AB are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, I. Further, at the corners A’, V and D, the incoming current I must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of I, using Kirchhoff’s first rule and the symmetry in the problem. Next take a closed loop, say, ABCC’EA, and apply Kirchhoff’s second rule:  IR  1/ 2IR  IR  ε  0 where R is the resistance of each edge and ε the emf of battery. Thus, ε  5 IR 2 The equivalent resistance of the network Req is : Req  ε  5 R 3I 6 For R = I Ω , Req  5 / 6Ω and ε = 10 V, the total current (= 3I) in the network is 3I = 10 V / (5/6) Ω = 12 A i.e., I = 4 A The current flowing in each edge can now be read off from the figure. It should be noted that because of the symmetry of the network, the great power of Kirchhoff’s rules has not been very apparent in Example. In a general network, there will be no such simplifica- tion due to symmetry, and only by application of Kirchhoff’s rules to junctions and closed loops (as many as necessary to solve the unknowns in the network) we can handle the problem. 12. In the circuit shown in figure, the battery is an ideal one with emf V. The capacitor is initially uncharged. The switch S is closed at time t = 0. (a) Find the charge Q on the capacitor at time t. (b) Find the current in AB at time t. What is its limiting value as t   ? B Solution: Let at any time ‘t’ charge on capacitor C be Q and current are as shown. Since charge Q will increase will time ‘t’ therefore. dQ N A i1 R S i1  dt (a) Applying Kirchoff’s second law in loop MNABM, V V  (i  i1)R  iR i i2=i–i1 R Q+ C R  V  2iR  i1R …(i) M B T Similarly, applying Kirchoff’s second law in loop MNSTM, V  i R  Q  iR …(ii) 1 C From equation (i) and (ii), V  i R  Q  V  i1R 1 C 2  V  3i R  2Q 1 C  3i R  V  2Q 1  i  C 1  V  2Q  1 3R  C     dQ  1  V  2Q  dt 3R  C    dQ  dt  V  2Q 3R C Q dQ t dt   0 V  2Q   dR C  Q  CV (1  e2 t / 3RC ) . Ans. 2 (b) From equation (i), V  V e2 t / 3 RC  i  V  i1R  3 2R 2R  Current through AB, i2  i  i1 V  V e2 t / 3 RC  3  V e2 t / 3RC 2R 3R i  V  V e2 t / 3RC 2 2R 6R As t   , i2  2R . 13. In the circuit shown in figure, R1  1, R 2  2 C1  1F , C2  2F and E  6V . Calculate charge on each capacitor in steady state. Solution: In steady state no current flow through capacitors. Therefore charge on each capacitor remains constant. Let, in steady state, the circuit draw a current I from the battery and let chare on capacitors be q1 and q2 as shown in the figure. Applying Kirchoff’s voltage law on mesh ABCFGHA, IR 2  E  IR1  0 I  E  2A or R 1  R 2 Now applying KVL on the mesh ABJHA, q1  IR  0 C1 or q1  2C (Negative sign indicates that the polarity of charge on capacitor C1 is opposite to assumed polarity. It means upper plate of the capacitor is negative while lower plate is positive. Hence, magnitude of charge on C1  2C Now applying KVL on mesh HJDFGH, q2  E  0 C2 or q 2  C2 E  12C . 60V 10 14. A two way switch S is used in the circuit shown in the figure. First S the capacitor is charged by putting the switch at position 1. Calcu- late heat generated across each resistor when the switch is shifted to position 2. Solution: Initially the switch was in position 1. Therefore, initially potential difference across capacitor was equal to the em.f. of the battery i.e., 60 V  initially energy stored in the capacitor was U  1 CV2 2  1  0.1 602 J  180J 2 q When switch is shifted to position 2 capacitor begins to discharge and energy stored in the capacitor is dissipated in the form of heat in the resistor. For a current I, applying Kirchoff’s law i1  i2  i And 6i1  3i2  0 , or i2  2i1  i1  1 i 3 and i2  2 i 3 But heat generated per unit time in the resistance is i2 R  The heat generated across 4,6 and 3 resistances are in ratio 4i2 : 6i2 : 3i2  6 : 1 :2 Total heat generated in the circuit per unit time  P1  P2  P3  U  Heat generated across 4 is P1  120 J Heat generated across 6 is Heat generated across 3 is P2  20J P3  40J Since during discharging, no current flows through 10 , therefore heat generated across it is equal to zero. 15. In the network shown in fig., find I, I1 and the charge on the 7F conditions have been reached. capacitor after equilibrium 7 F B 4  6  C A I1 4 2  3  D 6V I E 6V Soln After equilibrium conditions have been reached, no current passes through the capacitor branch. Thus at this state, no current will flow in AB and DB branches. Therefore I = I1 = current in DC branch. thus by applying Kirchhoffs, second law for the mesh ADCEA, we get 2I1  3I1  6  6  0 or I1  0. No current passes in the loop ABCEA, therefore there is no potential fall across the resistance in AB and BC arms. Thus the potential difference of 6V is applied across the plates of the 7F capacitor.