In the circuit shown, $R_1 = 4\Omega , R_2 = R_3 = 15\Omega , R_4 = 30\Omega$ and $E = 10V.$ Calculate. The equivalent resistance of the circuit and the current in each resistor.
CBSE DELHI - SET 1 2011
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$R_2, R_3$ and $R_4$ are in parallel.
$\frac{1}{\text{R}_{234}} = \frac{1}{15} + \frac{1}{30} + \frac{1}{15} = \frac{2+1+2}{30} = \frac{5}{30}$
$\Rightarrow\text{R}_{234} = 6 \Omega$
Now $R_{234}$ is in series with $R_1,$ so $R_{eq} = 4 \Omega+ 6 \Omega= 10 \Omega$
$\therefore\text{I} = \frac{\text{E}}{\text{R}_{eq}} = \frac{10}{10}\text{A} = 1\text{A}$
$\therefore\text{I}_{1} = 1 \text{A}$
$\therefore$ Current through $\text{R}_{1} = 1 \text{A}$
$P.D.$ across $\text{R}_{1} = 4 \text{V}$
So, $P.D.$ across $R_{234} = 6 V$
$\therefore\text{I}_{2}\text{R}_{2} = \text{I}_{4}\text{R}_{4} = \text{I}_{3}\text{R}_{3} = 6 \text{V}$
$\text{I}_{2} = \frac{6}{15}\text{A} = 0.4 \text{A}$
$\text{I}_{3} = \frac{6}{15}\text{A} = 0.4\text{A}$
$\text{I}_{4} = \frac{6}{30}\text{A} = 0.2\text{A}.$
$\frac{1}{\text{R}_{234}} = \frac{1}{15} + \frac{1}{30} + \frac{1}{15} = \frac{2+1+2}{30} = \frac{5}{30}$
$\Rightarrow\text{R}_{234} = 6 \Omega$
Now $R_{234}$ is in series with $R_1,$ so $R_{eq} = 4 \Omega+ 6 \Omega= 10 \Omega$
$\therefore\text{I} = \frac{\text{E}}{\text{R}_{eq}} = \frac{10}{10}\text{A} = 1\text{A}$
$\therefore\text{I}_{1} = 1 \text{A}$
$\therefore$ Current through $\text{R}_{1} = 1 \text{A}$
$P.D.$ across $\text{R}_{1} = 4 \text{V}$
So, $P.D.$ across $R_{234} = 6 V$
$\therefore\text{I}_{2}\text{R}_{2} = \text{I}_{4}\text{R}_{4} = \text{I}_{3}\text{R}_{3} = 6 \text{V}$
$\text{I}_{2} = \frac{6}{15}\text{A} = 0.4 \text{A}$
$\text{I}_{3} = \frac{6}{15}\text{A} = 0.4\text{A}$
$\text{I}_{4} = \frac{6}{30}\text{A} = 0.2\text{A}.$
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