Question

A battery $E_A$ of $123V$ emf and internal resistance $0.4\Omega$ and a battery $E_B$ of $117V$ emf and internal resistance $0.6\Omega$ supplies a current to load resistance $R_L$ through four wires of resistors $R_A=1.2\Omega ,R_B=1\Omega ,R_C=1.4\Omega$ and $R_D=1.4\Omega$ as shown in figure. The value of load resistance for maximum load power is

• A. $1.5\Omega$
• B. $3\Omega$
• C. $4.5\Omega$
• D. $6\Omega$

Answer 1

The correct option is A $1.5\Omega$

Given circuit can be rearranged as
Now for cells in parallel, effective emf is given by
$E_{eq}=\frac{E_A{r}_B+E_B{r}_A}{r_A+r_B}=\frac{123\left(3\right)+117\left(3\right)}{3+3}=120V\text{effective internal resistance is given by}{R}_{eq}=\frac{r_A{r}_B}{r_A+r_B}=1.5\Omega$

For external resistance to draw maximum power, its value should be equal to internal resistance of the cell.
Hence, load resistance for max load power is $R_L=R_{eq}=1.5\Omega$