Question

A circuit consists of eight identical batteries and a resistor $R=0.8\Omega$. Each battery has an open circuit emf of $1.0\text{V}$ and internal resistance of $0.2\Omega$. The voltage difference across any of the battery is

• A. $0.5\text{V}$
• B. $1.0\text{V}$
• C. $0\text{V}$
• D. $2.0\text{V}$

Answer 1

The correct option is C $0\text{V}$

Circuit can be redrawn as:



Using $\text{KCL}$, currents are marked as shown in the circuit.

Applying $\text{KVL}$ in loop $1$,

$4-4\left(0.2\right)i_0-(i_0-i_1)0.8=0$

$4-1.6i_0+0.8i_1=0....\left(1\right)$

Applying $\text{KVL}$ in loop $\left(2\right)$,

$4-4\left(0.2\right)i_1+(i_0-i_1)0.8=0$

$4+0.8i_0-1.6i_1=0....\left(2\right)$

Solving the above equations $1$ and $2$,

$-1.6i_0+0.8i_1=0.8i_0-1.6i_1$

$\Rightarrow 2.4i_1=2.4i_0$

$\Rightarrow i_1=i_0$

i.e., Current through $0.8\Omega$ is zero and putting $i_1=i_0$ in equation $1$ we get,

$i_0=5\text{A}$

Lets, calculate voltage across any one battery,


We can write,

$V_B-0.2\left(5\right)+1=V_A$

$V_A-V_B=0$

Hence, option (c) is the correct answer.