A) 50\[\Omega \]
B) 40\[\Omega \]
C) 20 \[\Omega \]
D) 10\[\Omega \]
Correct Answer: C
Solution :
Given case is balanced Wheatstone bridge. As in the upper and lower arms, the resistances are in series combination, hence, the resistance in upper arm is given by \[{{R}_{U}}=20+20=40\,\Omega \] Similarly, equivalent resistances in lower arm is \[{{R}_{L}}=20+20=40\,\,\Omega \] Now equivalent resistance \[{{\text{R}}_{\text{L}}}\] and \[{{\text{R}}_{\text{U}}}\] are connected in parallel combination hence, the resultant resistance is given by \[{{R}_{PQ}}=\frac{{{R}_{L}}{{R}_{U}}}{{{R}_{L}}+{{R}_{U}}}=\frac{40\times 40}{40+40}=20\,\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
