Question 122 of 380

As shown in the figure, a network of resistors is connected to a battery of $24V$ with an internal resistance of $3\Omega$. The currents through the resistors ${R}_{4}$ and ${R}_{5}$ are ${I}_{4}$ and ${I}_{5}$ respectively. The values of ${I}_{4}$ and ${I}_{5}$ are:

A${I}_{4}=\dfrac{8}{5}A$ and ${I}_{5}=\dfrac{2}{5}A$B${I}_{4}=\dfrac{24}{5}A$ and ${I}_{5}=\dfrac{6}{5}A$C${I}_{4}=\dfrac{6}{5}A$ and ${I}_{5}=\dfrac{24}{5}A$D${I}_{4}=\dfrac{2}{5}A$ and ${I}_{5}=\dfrac{8}{5}A$

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Answer: (D) ${I}_{4}=\dfrac{2}{5}A$ and ${I}_{5}=\dfrac{8}{5}A$

Equivalent resistance of circuit
${R}_{eq}=3+\left(\dfrac{2\times 2}{2+2}\right)+2+\left(\dfrac{20\times 5}{20+5}\right)+2=12\Omega$
Current through battery $i=\dfrac{24}{12}=2A$
Now, ${I}_{4}=\dfrac{{R}_{5}}{{R}_{4}+{R}_{5}}\times 2=\dfrac{5}{20+5}\times 2=\dfrac{2}{5}A$
and ${I}_{5}=i-{I}_{4}=2-\dfrac{2}{5}=\dfrac{8}{5}A$