Question

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: $(24V,3\Omega )$

• A. $I_4=\frac{8}{5}A$ and $I_5=\frac{2}{5}A$
• B. $I_4=\frac{6}{5}A$ and $I_5=\frac{24}{5}A$
• C. $I_4=\frac{2}{5}A$ and $I_5=\frac{8}{5}A$
• D. $I_4=\frac{24}{5}A$ and $I_5=\frac{6}{5}A$

Answer 1

Answer: (C)

Equivalent resistance of circuit
$R_{eq}=3+1+2+4+2$
$=12\Omega$
Current through battery $i=\frac{24}{12}=2A$
$I_4=\frac{R_5}{R_4+R_5}\times 2=\frac{5}{20+5}\times 2=\frac{2}{5}A$
$I_5=2-\frac{2}{5}=\frac{8}{5}A$