In the circuit shown, charge on the 5 μ F capacitor is :

Question

In the circuit shown, charge on the 5 μ F capacitor is : (A) 5.45 μ C (B) 16.36 μ C (C) 10.90 μ C (D) 18.00 μ C

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1d839eb4aac1326a458285d1a530b730-.png" /> Now, using junction analysis We can say, q 1+ q 2+ q 3=0 2(x-6)+4(x-6)+5(x)=0 x=(36/11) q3=(36(5)/11)=(180/11) q3=16.36 μ C

Q. In the circuit shown, charge on the $5 μ F$ capacitor is :

$5.45 μ C$

$16.36 μ C$

$10.90 μ C$

$18.00 μ C$

Now, using junction analysis
We can say, $q_{1} + q_{2} + q_{3} = 0$
$2 (x - 6) + 4 (x - 6) + 5 (x) = 0$
$x = \frac{36}{11} q_{3} = \frac{36 ( 5 )}{11} = \frac{180}{11}$
$q_{3} = 16.36 μ C$

Q. In the circuit shown, charge on the 5μF capacitor is :

5.45μC

16.36μC

10.90μC

18.00μC