question_answer

The equivalent resistance between the points X and Y in the circuit shown is

A)  \[\frac{2hv}{3}\]                             

B)  \[30\mu A\]

C)  \[90\mu A\]                                     

D)  \[4mA\]

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 \[90\mu A\]                         ....(i) Applying Kirchhoffs second law in a second loop XABCDYX, we get \[4mA\]                 \[2mA\]                      ?..(ii) Applying Kirchhoffs law in a closed loop ABEFA, we get                 \[3.6mA\]                 \[1.8mA\]                 \[62V,\,\,2\Omega \]                      ??.(iii) Applying Kirchhoffs law in closed loop BCDEB \[63V,\,\,1\Omega \]                                 \[61V,\,\,1\Omega \]                                 \[64V,\,\,2\Omega \]                                 \[15min\]          ??.(iv) From Eqs. (iii), and (iv), we get                                 \[20\text{ }min\]                                 \[7.5min\] From Eq. (iii), we get                 \[25min\]                 \[\frac{\omega M}{M+m}\]                 \[\frac{\omega (M-2M)}{M+2m}\] \[\frac{\omega (M+2M)}{M}\]  \[\frac{\omega M}{M+2m}\]                 \[{{a}_{c}}={{k}^{2}}{{r}^{2}}{{t}^{2}},\]                 \[2mk{{r}^{2}}t\] From Eq. (i)                 \[mk{{r}^{2}}{{t}^{2}}\]                 \[m{{k}^{2}}{{r}^{2}}t\]        ..?(iv) From Eq. (ii)                 \[m{{k}^{2}}r{{t}^{2}}\]    ??(vi) From Eqs. (v) and (vi), we get                 \[1\text{ }km/h\]                 \[\text{3 }km/h\]