A) \[790\,\Omega \]
B) \[810\,\Omega \]
C) \[990\,\Omega \]
D) \[1000\,\Omega \]
Correct Answer: C
Solution :
Potential gradient \[x=\frac{V}{L}=\frac{e}{(R+{{R}_{h}}+r)}\cdot \frac{R}{L}\] Þ \[2.2\times {{10}^{-3}}=\frac{2.2}{(10+{{R}_{h}})}\times 1\] Þ \[{R}'=990\,\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
