Title 18. Current Electricity Hard Ans.pdf ✅

1. (b) m s neA i vd 5 10 / 3 10 1.6 10 10 24 10 3 23 19 4 3 − − − − =       = = 2. (b) By using disc rod rod disc rod disc . ; A A l l R R A l R =  =   2 2 3 2 2 3 disc (10 ) (0.3 10 ) 1 10 3 10 − − − −  =    R   –7 2.7 10 R W disc =  . 3. (c) By using A l R = .  l  A  2 6 3.14 3.14 10 1 3.14  r   = −   r = 10–3 m = 1 mm 4. (a) In case of streching R  l 2 So 2 = 2  0.1 = 0.2  =  l l R R 5. (b) By using Rt = Ro (1 + t)  2 1 2 1 1 1 t t R R   + + = So 1 2 1 (300 273) 2 1  t  + + − =  t2 = 854oC = 1127 K 6. (b) V = constant., i = constant. So R = constant  2 2 Cu Cu Cu i i i Cu Cu Cu i i i r l r l A l A P l    =  =  2.4 17 100 1.7 10 1.0 10 8 7 =    = = − − Cu i Cu i r r   7. (d)     m l V l A l R 2 2 = = =       = V m  3 2 3 2 2 2 1 2 1 1 2 3 : : : : m l m l m l R R R = 125 : 15 : 1 5 1 : 3 9 = 25 : = 8. (a) All the conductors have equal lengths. Area of cross-section of A is   2 2 2 ( 3 a) − ( 2 a) = a Similarly area of cross-section of B = Area of cross-section of C = a 2 Hence according to formula ; A l R =  resistances of all the conductors are equal i.e. RA = RB = RC 9. (a) R = 10  101  20%  100  10. (a) P.d. between X and Y is VXY = VX – VY = 1  4 = 4 V. (i) and p.d. between X and Z is VXZ = VX – VZ = 1  16 = 16 V .... (ii) On solving equations (i) and (ii) we get potential difference between Y and Z i.e., reading of voltmeter is VY − VZ = 12V 11. (a) Initially : Resistance of given cable 3 2 (9 10 ) −   =   l R ..... (i) Finally : Resistance of each insulated copper wire is 3 2 (3 10 ) ' −   =   l R Hence equivalent resistance of cable           = =  −3 2 6 (3 10 ) 1 6 '   R l Req ..... (ii) On solving equation (i) and (ii) we get Req = 7.5  12. (d) Series resistance RS = R1 + R2 and parallel resistance 1 2 1 2 R R R R RP + =  n R R R R R R P S = + = 1 2 2 1 2 ( )  n R R R R 1 2 1 2 = +  n R R R R R R 1 2 2 2 1 2 2 1 + =  n R R R R 1 2 2 1 + = 13. (a) The equivalent circuit of above can be drawn as Which is a balanced wheatstone bridge. So current through AB is zero. So 10 1 20 1 20 1 1 = + = R  R = 10  14. (c) X 10  B A Y 10  10  10  20  l  l  9 mm 16  4  4  16  V 2A Y Z X 1A 1A
The equivalent circuit of above fig between A and D can be drawn as So Req = 10 + 10 + 10 = 30 15. (b) Taking the portion COD is figure to outside the triangle (left), the above circuit will be now as resistance of each is 2  the circuit will behaves as a balanced wheatstone bridge and no current flows through CD. Hence RAB = 2 16. (d) In the circuit, by means of symmetry the point C is at zero potential. So the equivalent circuit can be drawn as r r r Req 7 8 || 2 3 8  =      = 17. (a) Given Wheatstone bridge is balanced because S R Q P = . Hence the circuit can be redrawn as follows 18. (b) The potential difference across B, D will be zero, when the circuit will act as a balanced wheatstone bridge and 1 / 2 12 4 1 + 3 = + =  S x R Q P  x = 2 19. (b) In the given circuit 2A current divides equally at junction D along the paths DAC and DBC (each path carry 1A current). Potential difference between D and A, VD – VA = 1  2 = 2 volt .... (i) Potential difference between D and B, VD – VB = 1  3 = 3 volt ..... (ii) On solving (i) and (ii) VA – VB = + 1 volt 20. (d) On Solving further we get equivalent resistance is  3 8 On Solving further we get equivalent resistance is  3 8 21. (b) For zero deflection in galvanometer the potential different across X should be E = 2V In this condition 2 500 12 = + X X  X = 100  22. (b) The equivalent circuit can be drawn as since E1 & E3 are parallely connected So current i 2Amp 2 2 2 = + = from A to B. 23. (d) Current i 1A 3 2 1 10 4 = + + − = from a to b via e 24. (a) The given part of a closed circuit can be redrawn as follows. It should be remember that product of current and resistance can be treated as an imaginary cell having emf = iR.  7 Hence VA – VB = +9 V  7 Hence VA – VB = +9 V 25. (a) E1 = x (0.6) and E2 = E1 – 0.1 = x (0.55)  0.55 0.6 1 0.1 1 = E − E or 55 E1 = 60 E1 – 6  E 1.2 V 5 6 1 = = thus E2 = 1.1 V 26. (b) For loop (1) −2i1 − 2(i1 − i2 ) + 4 = 0  2i1 − i2 = 2 ... (i) For loop (2) 4 2( ) 6 0 − i2 + i1 − i2 + =  3 3 i2 − i1 = ... (ii) After solving equation (i) and (ii) we get i1 = 1.8 A and i2 = 1.6 A 4 V 3 V 2 V A B  9 V A B + – R = (R1 ||R2) = 2 A B 2V 2V 4 4 4 Series 4 + 4 = 8  Series 6 + 8 = 14  3  4  8  6  A B Series 3 + 4 = 7   7  14  A B Parallel =  3 14 Req r r r r r r r A B Series (r S r) = 2r  r r 2r r r r A B Parall el Serie  s r 3 2 r 2r r A B Series  A 2r B r r r r 3 8 3 2 + + =  C D E A B O  10 A 10 10 10 10 10 D Balanced wheatstone bridge 10 A 10 10 D Series
27. (a) Applying KVL in the given circuit we get −2i + 10 − 5 − 3i = 0  i = 1A Second method : Similar plates of the two batteries are connected together, so the net emf = 10 – 5 = 5V Total resistance in the circuit = 2 + 3 = 5  A R V i 1 5 5 = =   = 28. (a) The current in the circuit are assumed as shown in the fig. Applying KVL along the loop ABDA, we get – 6i1 – 3 i2 + 15 = 0 or 2i1 + i2 = 5 ...... (i) Applying KVL along the loop BCDB, we get – 3(i1 – i2) – 30 + 3i2 = 0 or – i1 + 2i2 = 10 ...... (ii) Solving equation (i) and (ii) for i2, we get i2 = 5 A 29. (c) i = 15 + 3 + 5 = 23 A 30. (d) Suppose current through different paths of the circuit is as follows. After applying KVL for loop (1) and loop (2) We get 28i1 = −6 − 8  i A 2 1 1 = − and 54i2 = −6 − 12  i A 3 1 2 = − Hence i i i A 6 5 3 = 1 + 2 = − 31. (a) 2 1 r x  Rp   9 1 3 2 2 1 2 2 2 1  =      = = r r r r x x 32. (c) By using KVL −5  2 − VPQ + 15 = 0  VPQ = 5V 33. (c) Let resistance of voltmeter be RV. Equivalent resistance between X and Y is V V XY R R R + = 75 75 Reading of voltmeter = potential difference across X and Y = 120 = i  RXY = V V R R +  75 75 2  RV = 300 34. (a) Ammeter has no resistance so there will be no potential difference across it, hence reading of voltmeter is zero. 35. (d) Resistance of voltmeter V1 is R1 = 200  80 = 16000  and resistance of voltmeter V2 is R2 = 32000  By using relation ; ' ' V R R V eq         = where V = potential difference across any resistance R in a series grouping. So for voltmeter V1 potential difference across it is V R R R 80 . 1 2 1         + =  V = 240 V 36. (c) By using i S i g 4 = 1 +  S 0.018 1 1 10 = +  S = 0.002  37. (a) The condition of wheatstone bridge gives , 80 20 r r R X = r- resistance of wire per cm, X- unknown resistance  =  =  1 = 0.25  4 1 80 20 X R 38. (b) Fraction of current passing through the galvanometer S G S i ig + = or 0.2 2 8 2 = + = i ig So fraction of current passing through the shunt amp i i i i s g = 1 − = 1 − 0.2 = 0.8 39. (a) Potential gradient A i AL i L L iR L V   = = = = X R = 1 P = 20 r Q = 80 r 20 cm 80 cm V1 R2 80 V V V2 R1 + – 12 V 54 i3 28 6 V 8 V 1 2 6 A C 15 V 3 B D 30 V i1 3 i1 – i2 i2 i1 2 i2 i1 i2 i1 (i1 – i2) 2 4 4V 6V 1 2
10 V/ m 8 10 0.2 40 10 2 6 8 − − − =    = 40. (b) By using  = =  −   = − = 0.5 2 1 2 100 125 100 ' 2 1 2 R r l l l r 41. (a) By using L R R R r e x h . ( + + ) =  10 30 (30 5) 2.5 10 50 10 3 6  + + =  − − R  R = 115 42. (a) By using L R R R r e x h . ( + + ) =  x 0.5 V/m 0.005 V/cm 1 5 (5 15 0) 2  = = + + = 43. (b) By using ' 2 1 2 R l l l r         − =  5 2 1 2         − = l r ...... (i) and 10 3 1 3         − = l r ....... (ii) On solving (i) and (ii) r = 10  44. (b) By using l L R R R r e E h eq  + + = . ( )  3 5 5 (5 4 1) 10   + + E =  E = 3 volt 45. (d) If suppose emf’s of the cells are E1 and E2 respectively then E1 + E2 = x  6 ....... (i)[x = potential gradient) ] and E1 – E2 = x  2 .......(ii)  1 3 1 2 1 2 = − + E E E E  1 2 2 1 = E E 46. (a) 5 1 10 2 10 1 10 1 1 = + = = R or R1 = 5  R2 = 4, l1 = 40 cm, l2 = ? 1 2 2 1 R R l = l or l 32 cm 5 40 4 2 =  = 47. (a) Rg × 5 I =       − 5 I I × 4  Rg = 16  Case II G 2 I I1 Rg = 16 4 16 I1 = 6 4 2 (I–I1)  I1 = I/13 So decrease in current to previous current 13 8 I/ 5 I/ 5 I/13 = − = 48. (a) 2  2V 1  1V Q P Enet = 1 2 1 2 2 1 r r E r –E r + = 2 1 2 – 2 + = 0 49. (d) i = 7 7V = 1 A Current flows in anticlockwise direction in the loop. Therefore 0 – 1 × 2 – 1 × 2 – 5 = V1 V1 = – 9V 50. (b) 60 R V V1 V2 2 1 V V = R 60 ........(1) V1 + V2 = V .........(2) from (1) & (2)  V2 = 60 R VR + ...........(3) R V 2 2 = 60 watt ........(4) from (3) & (4) V2R = 60 (60 + R)2 V2R = 60(3600 + R2 + 60R) 0 = 60R2 + 3600 R – V2R + 60 × 3600 (3600 – V2 – 60 × 60 × 2) (3600 – V2 + 60 × 60 × 2) > 0 (3 × 3600 – V2 ) (–3600 – V2 ) > 0 3 × 3600 – V2 < 0 60 × 3600 < V2 60 × 3 < V 51. (d) Current through 20W = 20 6 current through 10W = 10 6 Total current supplied by voltage source