question_answer

In the circuit, E, F, G and H are cells of emf 2, 1, 3 and 1 V respectively. 2, 1, 3 and 1 \[\Omega \] are their respectives potential resistances. The potenitial difference between the nodes B and D is

A)  \[\frac{2}{13}V\]                            

B)  \[\frac{3}{13}V\]

C)  \[\frac{3}{13}V\]                            

D)  \[\frac{4}{13}V\]

E)  \[\frac{1}{10}V\]

Correct Answer: E

Solution :

                The circuit can be redrawn as Applying KVL in loop BADB, we get \[(2\,\,\Omega )\,{{i}_{1}}-2V+1\,V+(1\,\,\Omega ){{i}_{1}}+(2\,\,\Omega )\,({{i}_{1}}-{{i}_{2}})=0\] \[\Rightarrow \]               \[(5\,\Omega )\,{{i}_{1}}-(2\,\Omega )\,{{i}_{2}}=1\,V\]                                ?(i) Now, applying KVL in loop DCBD, we get \[-\,3\,V+(3\,\Omega ){{i}_{2}}+(1\,\,\Omega ){{i}_{2}}+1\,V-(2\,\Omega )\,({{i}_{1}}-{{i}_{2}})=0\] \[\Rightarrow \]               \[-\,(2\,\Omega ){{i}_{1}}+(6\,\Omega ){{i}_{2}}=2\,V\]                ?(ii) From Eqs. (i) and (ii), we get \[{{i}_{1}}=\frac{5}{13}A,\,\]\[{{i}_{2}}=\frac{6}{13}A\] So that \[{{i}_{1}}-{{i}_{2}}=\frac{-1}{13}A\] Also \[{{V}_{B}}-{{V}_{D}}=(2\,\Omega )\,\left( \frac{1}{13}A \right)=2/13\,V\]