Question

Two cells are connected to an external load of resistance $2\Omega$ as shown in figure. The current through the load resistance will be :

• A. $2\text{A}$
• B. $1.5\text{A}$
• C. $3\text{A}$
• D. $4\text{A}$

Answer 1

The correct option is A $2\text{A}$

From the given circuit, we can observe that one non-ideal battery $(E_1=6\text{V},r_1=1\Omega )$ and one ideal battery $(E_2=4\text{V},r_1=0\Omega )$ are connected in parallel. So, net emf will be

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

Here both the batteries are connected in same polarity, hence $+ve$ sign has been taken.

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

$\Rightarrow E_{eq}=\frac{(6\times 0)+(4\times 1)}{1+0}=4\text{V}$

Equivalent internal resistance,

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

$\Rightarrow r_{eq}=\frac{0\times 1}{0+1}=0\Omega$

The equivalent circuit can be drawn as:


So, current in circuit will be

$i=\frac{E_{eq}}{R}=\frac{4}{2}=2\text{A}$

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

Why this question ? Tip : For parallel combination of battery we use $-ve$ sign in the formula for the reverse polarity of battery.