Four capacitors of capacitance 10, mu, F and | vedclass

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Question

When switch is open $=C_{\mathrm{eq} .}=\frac{15}{2}\, \mu \mathrm{F}$

$\mathrm{q}_{\mathrm{i}}=\mathrm{C}_{\mathrm{eq}} \mathrm{V}=\frac{15}{2}

Vedclass · Accepted Answer

$4.5 	imes 10^{-3}\,C$

Vedclass · Answer

When switch is open $=C_{\mathrm{eq} .}=\frac{15}{2}\, \mu \mathrm{F}$

$\mathrm{q}_{\mathrm{i}}=\mathrm{C}_{\mathrm{eq}} \mathrm{V}=\frac{15}{2} 	imes 200=1500 \,\mu \mathrm{C}$

When switch is closed $\mathrm{C}_{\mathrm{eq}}=30\, \mu \mathrm{F}$

$\mathrm{q}_{\mathrm{f}}=30 	imes 200=6000\, \mu \mathrm{C}$

Charge flow through

$\mathrm{AB}=\mathrm{q}_{\mathrm{f}}-\mathrm{q}_{\mathrm{i}}=4500\, \mu \mathrm{C}=4.5 	imes 10^{-3} \mathrm{\,C}$

Four capacitors of capacitance $10\, \mu\, F$ and a battery of $200\,V$ are arranged as shown. How much charge will flow through $AB$ after the switch $S$ is closed?

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