Question

If $V_{AB}=4\text{V}$ in the given figure, then the value of internal resistance $x$ of battery having emf $2\text{V}$ will be

• A. $5\Omega$
• B. $10\Omega$
• C. $15\Omega$
• D. $20\Omega$

Answer 1

The correct option is D $20\Omega$

The given circuit can be visualized as a combination of two batteries in parallel.

The emf & internal resistance of batteries are,

$E_1=5\text{V},r_1=10\Omega ,E_2=2\text{V},r_2=x\Omega$

Since, both batteries are in parallel combination with same polarity, so net emf will be

$E_{eq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$

$\Rightarrow E_{eq}=\frac{(5\times x)+(2\times 10)}{10+x}=\frac{5x+20}{x+10}\text{V}$

Equivalent internal resistance,

$r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$

$\Rightarrow r_{eq}=\frac{10\times x}{10+x}=\frac{10x}{10+x}\Omega$

The equivalent circuit can be drawn as:


Apply KVL between points $\text{A}$ and $\text{B}$,

$V_A+E_{eq}-i{r}_{eq}=V_B$

since, circuit is open, so current, $i=0$

$\Rightarrow E_{eq}=V_B-V_A$

Given that $V_B-V_A=4\text{V}$

$\Rightarrow \frac{5x+20}{x+10}=4$

$\Rightarrow 40+4x=5x+20$

$\therefore x=20\Omega$

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

Why this question ? Tip : If there would across have been a load resistance connected across points $\text{A}$ and $\text{B}$ in the given arrangement, then current would flow in equivalent circuit, hence in that scenario the potential difference across points $\text{A}$ and $\text{B}$ will not be equal to $E_{eq}$ i.e., $V_{AB}\ne E_{eq}$.