A network of five capacitors is shown in the figure. The total charge stored in the network connected between A and B is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10 μC
Let top node = $v_1$, bottom node = $v_2$, $V_A=3\,$V, $V_B=0\,$V.
Node equations:
$4(3-v_1)=2(v_1-0)+5(v_1-v_2)$
$6(3-v_2)=3(v_2-0)+5(v_2-v_1)$
Simplify:
$-11v_1+5v_2=-12$
$5v_1-14v_2=-18$
Solving gives $v_1=2\,$V and $v_2=2\,$V.
Charge drawn from $A$: $Q= C_1(3-v_1)+C_3(3-v_2)=4(1)+6(1)=10\ \mu\text{C}$.
Answer: $10\ \mu\text{C}$