Question

A $5V$ battery with internal resistance $2\Omega$ and a $2V$ battery with internal resistance $1\Omega$ are connected to a $10\Omega$ resistor as shown in figure, the current in $10\Omega$ resistor is?

• A. $0.27$A
• B. $0.31$A
• C. $0.031$A
• D. $0.53$A

Answer 1

The correct option is C $0.031$A

$\text{Step 1: Assuming Batteries to be in Parallel [Refer Figure 1]}$

The given circuit can be redrawn as in Figure 1.

We can assume both batteries to be in parallel connected to $10\Omega$ resistance.

In parallel battery combination:

$\frac{E_{eq}}{r_{eq}}=\frac{E_1}{r_1}+\frac{E_2}{r_2}$ $....\left(1\right)$

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

$=\frac{1}{2}+\frac{1}{1}=\frac{3}{2}$

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

Here the polarities are unlike, So $E_1=5V$ and $E_2=-2V$

So, Equation $\left(1\right)$ $\Rightarrow \frac{E_{eq}}{2/3}=\frac{5}{2}-\frac{2}{1}=0.5$

$\Rightarrow E_{eq}=\frac{1}{3}$

$\text{Step 1: Solving Equivalent circuit [Refer Figure 2]}$

$R_{eq}=r_{eq}+10\Omega =\frac{32}{3}\Omega$

Hence, Applying KVL on equivalent circuit, we get:

$i=\frac{E_{eq}}{R_{eq}}=\frac{1}{32}=0.031A$

Option $\text{C}$ is the correct answer

$\text{Alternate Solution:}$

$\bullet$ Assume currents in all three branches.

$\bullet$ Aply KCL at any one junction (Either $P_1$ or $P_2$) to find relation between the currents.

$\bullet$ Apply KVL in both left and right loops, and obtain the current in $10\Omega$ resistance.