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)

$I _4=\frac{2}{5} A$ and $I _5=\frac{8}{5} A$

Answer 2

The correct answer is (C) : $I _4=\frac{2}{5} A$ and $I _5=\frac{8}{5} A$
Equivalent resistance of circuit
Req​=3+1+2+4+2
=12Ω

Current through battery i=1224​=2A

I4​=R4​+R5​R5​​×2=20+55​×2=52​A

I5​=2−52​=58​A