Question

As shown in the figure, a network of resistors is connected to a battery of $24 V$ 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: $(24 V , 3 \Omega)$

• 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 $i =\frac{24}{12}=2 A$
$ 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$