In circuit shown below, the resistances are given in $ohms$ and the battery is assumed ideal with emf equal to $3\, volt$. The voltage across the resistance $R_4$ is ................. $V$
Diffcult
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(a) Equivalent resistance of the given network ${R_{eq}} = 75\,\Omega $
Total current through battery $i = \frac{3}{{75}}$
${i_1} = {i_2} = \frac{3}{{75 \times 2}} = \frac{3}{{150}}$
Current through ${R_4} = \frac{3}{{150}} \times \frac{{60}}{{(30 + 60)}}$$ = \frac{3}{{150}} \times \frac{{60}}{{90}} = \frac{2}{{150}}\,A$
${V_4} = {i_4} \times {R_4} = \frac{2}{{150}} \times 30 = \frac{2}{5}V = 0.4\,V$
Total current through battery $i = \frac{3}{{75}}$
${i_1} = {i_2} = \frac{3}{{75 \times 2}} = \frac{3}{{150}}$
Current through ${R_4} = \frac{3}{{150}} \times \frac{{60}}{{(30 + 60)}}$$ = \frac{3}{{150}} \times \frac{{60}}{{90}} = \frac{2}{{150}}\,A$
${V_4} = {i_4} \times {R_4} = \frac{2}{{150}} \times 30 = \frac{2}{5}V = 0.4\,V$
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