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

Question

Two batteries of emf 4 V and 8 V with internal resistance 1 Ω and 2 Ω are connected in a circuit with resistance of 9 Ω as shown in figure. The current and potential difference between the points P and Q are (A) (1/3) A and 3 V (B) (1/6) A and 4 V (C) (1/9) A and 9 V (D) (1/12) A and 12 V

Tardigrade Admin · Accepted Answer

I=(8-4/1+2+9)=(4/12)=(1/3) A VP-VQ=4-(1/3) × 3=3 Volt

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

$\frac{1}{3}$ A and 3 V

$\frac{1}{6}$ A and 4 V

$\frac{1}{9}$ A and 9 V

$\frac{1}{12}$ A and 12 V

$I = \frac{8 - 4}{1 + 2 + 9} = \frac{4}{12} = \frac{1}{3} A$
$V_{P} - V_{Q} = 4 - \frac{1}{3} \times 3 = 3$ Volt

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

31​ A and 3 V

61​ A and 4 V

91​ A and 9 V

121​ A and 12 V