In the circuit shown- charge on the 5 -F capacitor is

Let q_1 and q_2 be the charge on the capacitors of nbsp; 2μ F and 4μ F. Then charge on capacitor of 5μ F Q = q_1+q_2 ⇒ 5V_0=2(6 V_0) +4(6 V_0) ⇒ 5V_0=12 2V ...

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Byju's Answer

Standard XII

Physics

New Capacitance in Dielectric

In the circui...

Question

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

18.00 μ C

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10.90 μ C

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5.45 μ C

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16.36 μ C

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Solution

The correct option is D $16.36 μ C$

Let $q_{1}$ and $q_{2}$ be the charge on the capacitors of $2 μ F$ and $4 μ F$ . Then charge on capacitor of $5 μ F$

$Q = q_{1} + q_{2}$

$\Rightarrow 5 V_{0} = 2 (6 - V_{0}) + 4 (6 - V_{0})$
$\Rightarrow 5 V_{0} = 12 - 2 V_{0} + 24 - 4 V_{0}$
$\Rightarrow 11 V_{0} = 36 \Rightarrow V_{0} = \frac{36}{11} V$
$\Rightarrow Q = 5 V_{0} = \frac{180}{11} = 16.36 μ C$

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In the circuit shown, charge on the 5μF capacitor is

The correct option is D 16.36 μC Let q1 and q2 be the charge on the capacitors of 2μF and 4μF. Then charge on capacitor of 5μF Q=q1+q2 ⇒5V0=2(6−V0)+4(6−V0) ⇒5V0=12−2V0+24−4V0 ⇒11V0=36⇒V0=3611V ⇒Q=5V0=18011=16.36 μC

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