A) \[80\,\Omega \]
B) \[100\,\Omega \]
C) \[60\,\Omega \]
D) \[120\,\Omega \]
Correct Answer: B
Solution :
[b] |
\[R\to \]Resistance |
Potential gradient for the potentiometer wire |
AB is \[-\frac{dV}{d\ell }=\left[ \frac{60\times R}{{{\ell }_{AB}}} \right]mv/m\] |
\[\therefore {{V}_{AP}}=\left( -\frac{dV}{d\ell } \right){{\ell }_{AP}}=\frac{60\times R}{1200}\times 1000mV\] |
\[\therefore {{V}_{AP}}=50RmV\] |
Also, \[{{V}_{AP}}=5\text{ }V\](for balance point at P) |
\[\therefore R=\frac{5}{50\times {{10}^{-3}}}=100\Omega \] |
You need to login to perform this action.
You will be redirected in
3 sec