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

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