Question

Two batteries A and B and three resistors are connected in a circuit. Internal resistance of both the batteries is $1\Omega$ as shown. EMF of battery B is $5\text{V}$. The potential difference between P and Q is zero. Which of following is/are true ?

• A. The current through $5\Omega$ is $3\text{A}$.
• B. The current through the battery A is $3\text{A}$.
• C. The emf of the source A is $47\text{V}$.
• D. The potential difference between O and P is $8\text{V}$.

Answer 1

Answer: (A), (C)

The emf of the source A is $47\text{V}$.

Assume the direction of currents in the branches $I_1,I_2$ & $I_3$ as shown.


Given, potential difference between P and Q is zero.
$V_{PQ}=0$
Using KVL,
$5-I_1\times 1=0$
$\Rightarrow I_1=5\text{A}$ (current across PQ)

So, voltage drop across the $3\Omega$ resistance
$V_x=-5\times 3$
$V_x=-15\text{V}$

Since $V_P=V_Q$, voltage drop across $5\Omega =-15\text{V}$
$\Rightarrow I_3\times 5\Omega =15\text{V}$
$\Rightarrow I_3=3\text{A}$

Using KCL at point P, $I_2=I_1+I_3$
$=5+3=8\text{A}$

Voltage drop across left $3\Omega$ resistor $=8\times 3=24\text{V}$
or $V_O=24\text{V}$
$\Rightarrow V_O-V_P=24-(-15)=39\text{V}$

But, we know $V_O-V_P=E-I_2\left(1\Omega \right)$
$\Rightarrow 39=E-8\times 1$
$\Rightarrow E=39+8$
or $E=47\text{V}$

Hence, options (a) and (c) are correct.