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A $6 \mathrm{~V}$ cell with $0.5 \Omega$ internal resistance, a $10 \mathrm{~V}$ cell with $1 \Omega$ internal resistance and a $12 \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \mathrm{~V}$ cell is
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Verified Answer
The correct answer is:
2.87
In closed loop $A B G F E H A$,
$\begin{aligned}
10-i_2 \times 1+i_1 \times 0.5-6 & =0 \\
0.5 i_1-i_2 & =-4
\end{aligned}$
In closed loop $B C D E B$,
$\begin{array}{r}
\left(i_1+i_2\right) \times 12+i_2 \times 1-10=0 \\
12 i_1+13 i_2=10
\end{array}$
From Eqs. (i) and (ii), we get
$i_2=2.87 \mathrm{~A}$
$\begin{aligned}
10-i_2 \times 1+i_1 \times 0.5-6 & =0 \\
0.5 i_1-i_2 & =-4
\end{aligned}$
In closed loop $B C D E B$,
$\begin{array}{r}
\left(i_1+i_2\right) \times 12+i_2 \times 1-10=0 \\
12 i_1+13 i_2=10
\end{array}$
From Eqs. (i) and (ii), we get
$i_2=2.87 \mathrm{~A}$
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