In the given figure, charge stored in capacitor in steady state will be

Question

In the given figure, charge stored in capacitor in steady state will be (A) 32 μ C (B) 16 μ C (C) 24 μ C (D) 8 μ C

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/eb6f320bcdfdfbfcc4e8e9c3f893a0fc-.png" /> In steady state current drawn from battery I=(12/4+2)=2 A Potential drop across A B=4 × 2=8 V This is drop across capacitor also q=C V=2 × 10-6 × 8=16 μ C

Q. In the given figure, charge stored in capacitor in steady state will be

$32 μ C$

$16 μ C$

$24 μ C$

$8 μ C$

In steady state current drawn from battery
$I = \frac{12}{4 + 2} = 2 A$
Potential drop across $A B = 4 \times 2 = 8 V$
This is drop across capacitor also
$q = C V = 2 \times 1 0^{- 6} \times 8 = 16 μ C$

Q. In the given figure, charge stored in capacitor in steady state will be

32μC

16μC

24μC

8μC

In steady state current drawn from battery I=4+212​=2A Potential drop across AB=4×2=8V This is drop across capacitor also q=CV=2×10−6×8=16μC

$32 μ C$

$16 μ C$

$24 μ C$

$8 μ C$

In steady state current drawn from battery
$I = \frac{12}{4 + 2} = 2 A$
Potential drop across $A B = 4 \times 2 = 8 V$
This is drop across capacitor also
$q = C V = 2 \times 1 0^{- 6} \times 8 = 16 μ C$