Two batteries of emf 3 V and 6 V with internal resistances 2 and 4 are connected in a circuit with resistance of 10 as shown in figure. The current and potential difference between the points P and Q are

Correct answer: D. (3)/(16) ~A and (15)/(8) ~V Applying Kirchhoff's voltage law in the given loop, aligned -4 I+6-3-2 I-10 I & =0 -16 I & =-3 I & =(3)/(16) ~A aligned Potential difference across P Q=(3)/(16) × 10=(15)/(8) ~V

JIPMER 2016

Exam: JIPMER
Year: 2016
Subject: Physics
Chapter: Current Electricity

JIPMER 2016 · Physics · Current Electricity

Two batteries of emf 3 V and 6 V with internal resistances 2 and 4...

Two batteries of emf 3 V and 6 V with internal resistances

2 \Omega

and

4 \Omega

are connected in a circuit with resistance of

10 \Omega

as shown in figure. The current and potential difference between the points

P

and

Q

are

Two batteries of emf 3 V and 6 V with internal resistances 2 and 4 are connected in a circuit with resistance of 10 as

Options

A $\frac{3}{16} \mathrm{~A}$ and $\frac{8}{15} \mathrm{~V}$
B $\frac{16}{3} \mathrm{~A}$ and $\frac{15}{8} \mathrm{~V}$
C $\frac{3}{16} \mathrm{~A}$ and 8 V
D $\frac{3}{16} \mathrm{~A}$ and $\frac{15}{8} \mathrm{~V}$

Correct answer

D. $\frac{3}{16} \mathrm{~A}$ and $\frac{15}{8} \mathrm{~V}$

Step-by-step solution

Applying Kirchhoff's voltage law in the given loop, Two batteries of emf 3 V and 6 V with internal resistances 2 and 4 are connected in a circuit with resistance of 10 as

\begin{aligned}-4 I+6-3-2 I-10 I & =0 \\ -16 I & =-3 \\ I & =\frac{3}{16} \mathrm{~A}\end{aligned}

\because

Potential difference across

P Q=\frac{3}{16} \times 10=\frac{15}{8} \mathrm{~V}$

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