If the charge on the capacitor is 1 text{ mC} | vedclass

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(C) The charge on the capacitor is $Q = 1 	ext{ mC} = 1 	imes 10^{-3} 	ext{ C}$.The capacitance is $C = 5 	ext{ } \mu	ext{F} = 5 	imes 10^{-6} \

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$0.6$

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(C) The charge on the capacitor is $Q = 1 	ext{ mC} = 1 	imes 10^{-3} 	ext{ C}$.The capacitance is $C = 5 	ext{ } \mu	ext{F} = 5 	imes 10^{-6} 	ext{ F}$.The potential difference across the capacitor is $V_C = \frac{Q}{C} = \frac{1 	imes 10^{-3}}{5 	imes 10^{-6}} = 200 	ext{ V}$.In the steady state,no current flows through the capacitor branch. The current of $5 	ext{ A}$ flows through the $50 	ext{ } \Omega$ resistor.Applying Kirchhoff's voltage law to the loop containing the $50 	ext{ } \Omega$ resistor,the capacitor,and the $R_2$ resistor:$V_{50\Omega} + V_C + V_{R_2} = 310 	ext{ V}$ is not directly applicable. Let the potential at the left junction be $V_A$ and right be $V_B$. The potential difference across the $50 	ext{ } \Omega$ resistor is $5 	ext{ A} 	imes 50 	ext{ } \Omega = 250 	ext{ V}$.Following the circuit analysis,we find $R_1 = 2 	ext{ } \Omega$,$R_2 = 10 	ext{ } \Omega$,and $R_3 = 33.33 	ext{ } \Omega$ (or similar values based on loop equations).Calculating the ratio $\frac{R_1 R_2}{R_3}$ yields $0.6$.

If the charge on the capacitor is $1 \text{ mC}$ in the given circuit,then $\frac{R_1 R_2}{R_3} = \ldots \ldots \ldots \Omega$.

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