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Question

In steady state there will be no current in branch of capacitor, so no voltage drop across $\mathrm{R}_2=5 \Omega$

$\mathrm{I}_2=0$

$\mathrm{I}_

Vedclass · Accepted Answer

$60$

Vedclass · Answer

In steady state there will be no current in branch of capacitor, so no voltage drop across $\mathrm{R}_2=5 \Omega$

$\mathrm{I}_2=0$

$\mathrm{I}_1=\mathrm{I}_3=\frac{10}{4+6}=1 \mathrm{~A}$

$\mathrm{~V}_{\mathrm{R}_3}=\mathrm{V}_{\mathrm{c}}+\mathrm{V}_{\mathrm{R}_2} \quad \mathrm{~V}_{\mathrm{R}_2}=0$

$\mathrm{I}_3 \mathrm{R}_3=\mathrm{V}_{\mathrm{c}}$

$\mathrm{V}_{\mathrm{c}}=1 	imes 6=6 \mathrm{volt}$

$\mathrm{q}_{\mathrm{c}}=\mathrm{CV}_{\mathrm{c}}=10 	imes 6=60 \mu \mathrm{C}$

In an electrical circuit drawn below the amount of charge stored in the capacitor is___ $\mu \mathrm{C}$.

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