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

Question

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

Aicharya · Accepted Answer

Charge on cap $=1 \mathrm{mC}$ $ \mathrm{V}_{\text {capacitor }}=\frac{Q}{C}=\frac{1 \times 10^{-3}}{5 \times 10^{-6}}=\frac{1000}{5}=200 \mathrm{~V} $ So, current through $R_5$ $ I=\frac{V}{R}=\frac{200}{10}=20 \mathrm{~A} $ Hence, we have following current distribution.

In loop $B C D B$,

In loop $A B C D A$, $ \begin{aligned} & -200-15 R_3+50+250=0 \\ & 15 R_3=100 \Rightarrow R_3=\frac{100}{15} \Omega \end{aligned} $ From Eq. (i), we get $ 5 R_2=50 \Rightarrow R_2=10 \Omega $ and from loop $A D C A$, $ \begin{aligned} & -250-50+310-25 R_1=0 \\ & 25 R_1=10 \Rightarrow R_1=\frac{10}{25} \Omega \end{aligned} $ So, $\frac{R_1 \times R_2}{R_3}=\frac{\frac{10}{25} \times 10}{\left(\frac{100}{15}\right)}=\frac{15}{25}=\frac{3}{5}=0.6$

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

Solution