question_answer

Two batteries of emf 4 V and 8 V having the internal resistance of \[1\,\Omega \] and \[2\,\Omega \] respectively are connected in circuit with a resistance of \[9\,\Omega \] as shown in a figure. The current and potential difference between the points P and Q are:

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

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

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

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

Correct Answer: D

Solution :

Key Idea: Kirchhoffs second law is applied w a closed mesh of an electric circuit. From Kirchhoffs second law in any closed mesh of a circuit the algebraic sum of the products of currents and resistance in each part of the mesh is equal to the algebraic sum of the emfs in that mesh that is, \[\Sigma \,iR=E\] \[8-i{{r}_{2}}=iR-i{{r}_{1}}-4=0\]                 Here,     \[R=9\,\Omega ,{{r}_{1}}=1\,\Omega ,\,{{r}_{2}}=2\,\Omega \]                 \[\therefore \]  \[8-i\times 2-i\times 9-i\times 1-4=0\]                 \[\Rightarrow \]               \[i=\frac{4}{12}=\frac{1}{3}A\] Potential difference across PQ is                 \[{{V}_{PQ}}=i\,R=\frac{1}{3}\times 9=3\,V\]