A) 4 ohm
B) 6 ohm
C) 8 ohm
D) 9 ohm
Correct Answer: C
Solution :
Potential difference between B and D is zero, it means Wheatstone bridge is in balanced condition So \[\frac{P}{Q}=\frac{R}{S}\] Þ \[\frac{21}{3+\frac{8X}{(8+X)}}=\frac{18}{6}\]Þ \[X=8\Omega \]You need to login to perform this action.
You will be redirected in
3 sec