Question

For the circuit shown $R_1=1\Omega ,R_2=2\Omega ,E_1=2\text{V}$ and $E_2=E_3=4\text{V}$. The potential difference between the points $a$ and $b$ is approximately (in $\text{Volts}$)

• A. $2.7$
• B. $2.3$
• C. $3.7$
• D. $3.3$

Answer 1

The correct option is D $3.3$

From the given circuit, we can observe that all three batteries are connected in a parallel connection across $a$ and $b$.

Also, the internal resistance for battery $E_1$ and $E_3$ can be given as $2R_1$.

Redrawing the circuit,


Applying the grouping relation for parallel combination of three non-ideal cells,

$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_1}{r_3}}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}}$


$E_{eq}=\frac{\frac{2}{2}+\frac{4}{2}+\frac{4}{2}}{\frac{1}{2}+\frac{1}{2}+\frac{1}{2}}$

$\therefore E_{eq}=\frac{10}{3}=3.3\text{V}$

Here, $\frac{1}{r_{eq}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}$

$\therefore r_{eq}=\frac{2}{3}\Omega$


There will be no current drawn from the equivalent cell because of open circuit.

$\therefore V_a-V_b=E_{eq}$

or, $V_a-V_b=3.33\text{V}$

Hence, option $\left(d\right)$ is correct.

Why this question ? Tip: whenever we find that some resistance and cell are connected in certain branches of an electric circuit, then it is recommended to apply grouping of cells to obtain an equivalent system.