question_answer

The equivalent resistance between points\[X\] and V in the circuit shown is

A)  \[10\,\Omega \]                             

B)  \[5\,\Omega \]

C)  \[7\,\Omega \]                               

D)  \[14\,\Omega \]

Correct Answer: C

Solution :

                 Let the distribution of current in the various arms as shown in figure. Let R is the effective resistance between\[X\]and Y then, \[E=(i+{{i}_{1}})R\]                   ...(i) Applying Kirchhoff?s second law in a second loop\[XABCDYX,\]we get                 \[-10i,5(i+{{i}_{2}})+E=0\] or          \[5{{i}_{1}}+5{{i}_{2}}=10i\]                                  ...(ii) Applying Kirchhoff's law in a closed loop ABEFA, we get \[-10i+5{{i}_{2}}+5{{i}_{1}}=0\] \[5{{i}_{1}}+5{{i}_{2}}=10i\] \[2i={{i}_{1}}-8{{i}_{2}}\]                                    ...(iii) Applying Kirchhoff?s law in closed loop BCDEB \[-5(i+{{i}_{2}})+10({{i}_{1}}-{{i}_{2}})-5{{i}_{2}}=0\]                 \[5i=10{{i}_{1}}-20{{i}_{2}}\]                 \[i=2{{i}_{1}}-4{{i}_{2}}\]                 \[2i=4{{i}_{1}}-8{{i}_{2}}\]                                       ...(iv) From Eqs. (iii) and (iv), we get                 \[{{i}_{1}}+{{i}_{2}}=4{{i}_{1}}-8{{i}_{2}}\]                 \[{{i}_{2}}=\frac{1}{3}{{i}_{1}}\] From Eq. (iii), we get \[2i={{i}_{1}}+\frac{1}{3}{{i}_{1}}\] \[2i=\frac{4}{3}{{i}_{1}}\] \[i=\frac{2}{3}{{i}_{1}}\Rightarrow {{i}_{1}}=\frac{3}{2}i\] \[\therefore \]  \[{{i}_{2}}=\frac{1}{3}\times \frac{3}{2}i\]                 \[{{i}_{2}}=\frac{1}{2}i\] From Eq. (i) \[E=(i+{{i}_{1}})R\] \[=\left( i+\frac{3}{2}i \right)R\] \[=\frac{5}{2}iR\]                                              ...(v) From Eq. (ii)                 \[E=15i+5\times \frac{1}{2}i\]                 \[=\frac{35}{2}i\]                                              ?..(vi) From Eqs. (v) and (vi), we get                 \[\frac{5}{2}iR=\frac{35}{2}i\]                 \[R=7\,\Omega \]