The charge on the 6 μ F capacitor in the circuit shown is

Question

The charge on the 6 μ F capacitor in the circuit shown is (A) 540 μ C (B) 270 μ C (C) 180 μ C (D) 90 μ C

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fad95637648d0b3d88bfa2c34393b841-.png" /> (1/C)=(1/18)+(1/9)=(3/18) C=6 μ F q=6 × 90=540 μ C (q1/540-q1)=(6/12) 2 q1=540-q1 3 q1=540 q1=180 μ C

Q. The charge on the $6 μ F$ capacitor in the circuit shown is

$540 μ C$

$270 μ C$

$180 μ C$

$90 μ C$

$\frac{1}{C} = \frac{1}{18} + \frac{1}{9} = \frac{3}{18}$
$C = 6 μ F$
$q = 6 \times 90 = 540 μ C$
$\frac{q _{1}}{540 - q _{1}} = \frac{6}{12}$
$2 q_{1} = 540 - q_{1}$
$3 q_{1} = 540$
$q_{1} = 180 μ C$

Q. The charge on the 6μF capacitor in the circuit shown is

540μC

270μC

180μC

90μC