Current Electricity - Result Question 39
41. Two batteries of emf $4 V$ and $8 V$ with internal resistance $1 \Omega$ 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 $P$ and $Q$ are
[1988]
(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) $\frac{1}{12} A$ and $12 V$
Topic 3: Kirchhoff’s Laws, Cells, Thermo emf & Electrolysis
Show Answer
Answer:
Correct Answer: 41. (a)
Solution:
- (a) $I=\frac{8-4}{1+2+9}=\frac{4}{12}=\frac{1}{3} A$;
$ V_P-V_Q=4-\frac{1}{3} \times 3=3 volt $
