The switch S has been closed for long time and the electric circuit shown carries a steady current. Let $C_{1} = 3 .0 μF, C_{2} = 6 .0 μF, R_{1} = 4 .0 kΩ, and R_{1} = 7 .0 kΩ .$ The power dissipated in $R_{2}$ is 2.8 W

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The charge on capacitor $C_{1}$ is $240 μ C$

Long time after switch is opened, the charge on $C_{1}$ is $660 μ C$

The power dissipated to the resistor $R_{1}$ is 1.6 W

The charge on capacitor $C_{2}$ is $440 μ C$

answer is A, B, D.

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Detailed Solution

$i = 2 \times 10^{- 2} A$

$P_{R_{1}} = i^{2} R_{1} = {(2 \times 10^{- 2})}^{2} \times 4 \times 10^{- 3} = 1.6 W$

$Q_{C_{1}} = V_{R_{1}} \times C_{1} = 80 \times 3 \times 10^{- 6} = 240 μ C$

$Q_{C_{2}} = V_{R_{2}} \times C_{2} = 140 \times 6 \times 10^{- 6} = 840 μ C$

$Q_{C_{1}} = E C_{1} = 220 \times 3 \times 10^{- 6} = 660 μ C$

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