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 \[\frac{5}{2}iR=\frac{35}{2}i\] \[R=7\,\Omega \]You need to login to perform this action.
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