Question

Figure shows four cells of emf $5\text{V}$ and internal resistance $5\Omega$ are connected in series. The voltage across each battery is

• A. $2\text{V}$
• B. $0\text{V}$
• C. $3\text{V}$
• D. $4\text{V}$

Answer 1

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


For the given circuit, effective emf will be

$E_0=5\times 4=20\text{V}$ [all cells are in same polarity]

And effective resistance of given series combination is given by

$R_0=5+5+5+5=20\Omega$

so, current in the circuit,

$I=\frac{E_0}{R_0}=\frac{25}{20}=1\text{A}$

Let us take the cell of arm $\text{AB}$, voltage across it,

$V=E-Ir$
$\Rightarrow V=5-(1\times 5)=0\text{V}$

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