Question

Two batteries of emf $4\text{V}$ and $8\text{V}$ with internal resistances $1\Omega$ and $2\Omega$ are connected in a circuit with a resistance of $9\Omega$ as shown in figure. The current and potential difference between the points $\text{P}$ and $\text{Q}$ are

• A. $\frac{1}{3}\text{A}$ and $3\text{V}$
• B. $\frac{1}{6}\text{A}$ and $4\text{V}$
• C. $\frac{1}{9}\text{A}$ and $9\text{V}$
• D. $\frac{1}{2}\text{A}$ and $12\text{V}$

Answer 1

The correct option is A $\frac{1}{3}\text{A}$ and $3\text{V}$


$E_1=4\text{V}$
$E_2=8\text{V}$

So, $E_2>E_1$ and current flows from Q to P.

Polarity of both the cells are opposite.

So, net emf applied in the circuit is, $8-4=4\text{V}$

All resistances are in series, Equivalent resistance,
$1+2+9=12\Omega$

$\therefore i=\frac{4}{12}=\frac{1}{3}\text{A}$

Now, Potential differences across

PQ $=i\times 9=\frac{1}{3}\times 9=3\text{V}$

Key concept: The Kirchhoff’s voltage law (KVL) states that the algebraic sum of the potential difference around any closed loop of an electric circuit is zero.