NEETPhysicsCapacitance
A network consisting of three resistors, three batteries, and a capacitor is shown in figure. ( array |l|l|l|l| & Column I & & Column II (A) & Current in branch EB & (P) & 10 , C (B) & Current in branch CB & (Q) & 0.5 , A (C) & Current in branch ED & (R) & 1.5 , A (D) & Charge on capacitor & (S) & 5 , C & & (T) & None of these array )
Options
- A( A R , B Q , C Q , D P )
- B( A R , B Q , C R , D S )
- C( A Q , B R , C R , D P )
- D( A R , B Q , C Q , D S )
Correct answer
A. ( A R , B Q , C Q , D P )
Step-by-step solution
When a steady state is reached, no current passes through the capacitor or the branch CE. Considering the loop ABEFA, (5 ( i ₁+ i ₂ )=10 or i ₁+ i ₂=2 ~A (i) ) Considering the loop BCDEB (4 i ₂=12-10=2 i ₂=0.5 ~A ) So ( i ₁=2-0.5=1.5 ~A ) To find the charge on capacitors, we must know potential difference across the plates. Consider the loop CEDC : (-12+4 i ₂+3 0- V _ C +8=0 ~V _ C =-2 ~V ) So charge on capacitor ( Q = CV =10 C )