Question

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

(a) \(\frac{3}{16}\)A and \(\frac{8}{15}\)V

(b) \(\frac{16}{3}\)A and \(\frac{15}{8}\)V

(c) \(\frac{3}{16}\)A and \(8\)V

(d) \(\frac{3}{16}\)A and \(\frac{15}{8}\)V

Answer 1

Answer is : (d) \(\frac{3}{16}\)A and \(\frac{15}{8}\)V

Applying Kirchhoff’s voltage law in the given loop.

- 4l + 6 - 3 - 2l - 10l = 0

- 16l = - 3

l = \(\frac{3}{16}\)A

∵ Potential difference across

PQ = \(\frac{3}{16}\) x 10

PQ = \(\frac{15}{8}\)V