First slide

Ohm's law

Question

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

Moderate

A

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

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

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

$\frac{1}{2}$ A and 12 V

Solution

Applying Kirchhoff's voltage law in the given loop,
$- 2 i + 8 - 4 - 1 \times i - 9 i = 0 \Rightarrow i = \frac{1}{3} A$
Potential difference across $PQ = \frac{1}{3} \times 9 = 3 V$

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