A part of circuit is shown in figure- Energy stored in 0-2 - F capacitor in steady state in microjoules is

Using junction law p d across a amp; b is V_ab = 3 × 2 + 5 nbsp;× 1 + 4 nbsp;× 1 = 15V By E = 1/2CV^2 = 1/2 nbsp;× 0.2 nbsp;× 10^ 6 nbsp;× 15^2 E= ...

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Standard XI

Physics

Energy of a Capacitor

A part of cir...

Question

A part of circuit is shown in figure. Energy stored in $0.2 μ F$ capacitor in steady state in microjoules is

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Solution

Using junction law p-d across a & b is
$V_{a b} = 3 \times 2 + 5 \times 1 + 4 \times 1 = 15 V$
By $E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 0.2 \times 10^{- 6} \times 15^{2} E = 22.5 \times 10^{- 6} J$

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A part of circuit is shown in figure. Energy stored in 0.2μF capacitor in steady state in microjoules is

Using junction law p-d across a & b is Vab=3×2+5×1+4×1=15V By E=12CV2=12×0.2×10−6×152E=22.5×10−6J

A part of circuit is shown in figure. Energy stored in $0.2 μ F$ capacitor in steady state in microjoules is

Using junction law p-d across a & b is
$V_{a b} = 3 \times 2 + 5 \times 1 + 4 \times 1 = 15 V$
By $E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 0.2 \times 10^{- 6} \times 15^{2} E = 22.5 \times 10^{- 6} J$