In the circuit shown, current through $R_2$ is zero. If $R_4 = 2\,\Omega $ and $R_3 = 4\,\Omega $ , current through $R_3$ will be ................. $\mathrm{A}$
Medium
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Current through $\mathrm{R}_{3}=\frac{15-(-5)}{4}=5 \mathrm{\,A}$
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