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In the given figure $\mathrm{R}_1=10 \Omega, \mathrm{R}_2=8 \Omega, \mathrm{R}_3=4 \Omega$ and $\mathrm{R}_4=8 \Omega$. Battery is ideal with emf $12 \mathrm{~V}$. Equivalent resistant of the circuit and current supplied by battery are respectively :
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Verified Answer
The correct answer is:
$12 \Omega$ and $1 \mathrm{~A}$
Here $\mathrm{R}_2, \mathrm{R}_3, \mathrm{R}_4$ are in parallel
$\begin{aligned}
& \frac{1}{\mathrm{R}_{234}}=\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}+\frac{1}{\mathrm{R}_4} \\
& \mathrm{R}_{234}=2 \Omega
\end{aligned}$
$R_{234}$ is in series with $R_1$ so
$\begin{aligned}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{234}+\mathrm{R}_1=2+10=12 \Omega \\
& \mathrm{i}=\frac{12}{12}=1 \mathrm{Amp}
\end{aligned}$
$\begin{aligned}
& \frac{1}{\mathrm{R}_{234}}=\frac{1}{\mathrm{R}_2}+\frac{1}{\mathrm{R}_3}+\frac{1}{\mathrm{R}_4} \\
& \mathrm{R}_{234}=2 \Omega
\end{aligned}$
$R_{234}$ is in series with $R_1$ so
$\begin{aligned}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{234}+\mathrm{R}_1=2+10=12 \Omega \\
& \mathrm{i}=\frac{12}{12}=1 \mathrm{Amp}
\end{aligned}$
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