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.

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 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