question_answer

Two batteries of e.m.f. 4V and 8 V with internal resistances 1 W and 2 W are connected in a circuit with a resistance of 9 W as shown in figure.  The current and potential difference between the points P and Q are                    [AFMC 1999]

A)            \[\frac{1}{3}A\,\,\text{and }\,3V\]

B)            \[\frac{1}{6}A\,\,\text{and}\,\text{ }4V\]

C)            \[\frac{1}{9}A\,\,\text{and}\,\text{ 9}V\]

D)            \[\frac{1}{2}A\,\,\text{and}\,\,12V\]

Correct Answer: A

Solution :

           Applying Kirchoff's voltage law in the given loop. \[-2i+8-4-1\times i-9i=0\] Þ \[i=\frac{1}{3}A\] Potential difference across PQ \[=\frac{1}{3}\times 9=3V\]