Question

For the circuit given the current through $4\Omega$ resistor and the potential difference $V_A-V_B$ respectively are:

• A. $1\text{A},4\text{V}$
• B. $2\text{A},3\text{V}$
• C. $2.5\text{A},5\text{V}$
• D. $1.5\text{A},2\text{V}$

Answer 1

The correct option is A $1\text{A},4\text{V}$

Using the relation for parallel combination of batteries,

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

Here, $E_1=12\text{V},E_2=6\text{V}$ & $r_1=3\Omega ,r_2=6\Omega$

$E_{eq}=\frac{\frac{12}{3}-\frac{6}{6}}{\frac{1}{3}+\frac{1}{6}}$

$\Rightarrow E_{eq}=\frac{4-1}{\frac{2+1}{6}}=\frac{3}{\left(1/2\right)}$

$\therefore E_{eq}=2\times 3=6\text{V}$

Equivalent internal resistance will be

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}$

$\Rightarrow \frac{1}{r}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}$

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

On redrawing the equivalent circuit, we get


So, equivalent resistance of the circuit will be

$R_{eq}=2+4=6\Omega$

So, current in the circuit will be

$i=\frac{E_{eq}}{R_{eq}}=\frac{6}{6}=1\text{A}$

From the original circuit we can infer that potential difference across $\text{A}$ and $\text{B}$ will be equal to potential difference across $4\Omega$ resistor.

Thus,

$V_A-V_B=iR$ (from Ohm's law)

$\Rightarrow V_A-V_B=1\times 4=4\text{V}$

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

Why this question ? Tip : In problems involving combined arrangement of cells & resistance, always simplify the circuit to equivalent cell. Therefore, identify the resistance and potential difference across it that can be obtained by using Ohm's law, if we find current in circuit.