Question
Two batteries of emf $$4\,V$$ and $$8\,V$$ with internal resistance $$1\,\Omega $$ and $$2\,\Omega $$ are connected in a circuit with a resistance of $$9\,\Omega $$ as shown in figure. The current and potential difference between the points $$P$$ and $$Q$$ are
Two batteries of emf $$4\,V$$ and $$8\,V$$ with internal resistance $$1\,\Omega $$ and $$2\,\Omega $$ are connected in a circuit with a resistance of $$9\,\Omega $$ as shown in figure. The current and potential difference between the points $$P$$ and $$Q$$ are
A.
$$\frac{1}{3}A$$ and $$3\,V$$
B.
$$\frac{1}{6}A$$ and $$4\,V$$
C.
$$\frac{1}{9}A$$ and $$9\,V$$
D.
$$\frac{1}{12}A$$ and $$12\,V$$
Answer :
$$\frac{1}{3}A$$ and $$3\,V$$
Solution :
$$\eqalign{ & I = \frac{{8 - 4}}{{1 + 2 + 9}} = \frac{4}{{12}} = \frac{1}{3}A; \cr & {V_P} - {V_Q} = 4 - \frac{1}{3} \times 3 = 3\,volt \cr} $$
$$\eqalign{ & I = \frac{{8 - 4}}{{1 + 2 + 9}} = \frac{4}{{12}} = \frac{1}{3}A; \cr & {V_P} - {V_Q} = 4 - \frac{1}{3} \times 3 = 3\,volt \cr} $$
