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In the given circuit, the current in $8 \Omega$ resistance is $1.5 \mathrm{~A}$. The total current (I) flowing in the circuit is
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Verified Answer
The correct answer is:
5.5 A
$$
\mathrm{I}_1=1.5 \mathrm{~A}, \mathrm{R}_1=8 \Omega
$$
P.D. across $\mathrm{R}_1=1.5 \times 8=12 \mathrm{~V}$
$R_2=3 \Omega ; R_2$ is in parallel with $R_1$ and hence P.D. across it is also $12 \mathrm{~V}$.
$$
\therefore \mathrm{I}_2=\frac{12}{3}=4 \mathrm{~A}
$$
Total current $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2=1.5+4=5.5 \mathrm{~A}$
\mathrm{I}_1=1.5 \mathrm{~A}, \mathrm{R}_1=8 \Omega
$$
P.D. across $\mathrm{R}_1=1.5 \times 8=12 \mathrm{~V}$
$R_2=3 \Omega ; R_2$ is in parallel with $R_1$ and hence P.D. across it is also $12 \mathrm{~V}$.
$$
\therefore \mathrm{I}_2=\frac{12}{3}=4 \mathrm{~A}
$$
Total current $\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2=1.5+4=5.5 \mathrm{~A}$
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