Saved Bookmarks
| 1. |
A wheatstone bridge is almost balanced, with point C grounded.Calculate :- (a) The potential of point B (b) The potential of point D (c) If a galvanometer is connected between B and D, what is the direction of current through it? (d) For what value of the resistance BC, would the bridge be in balanced state? |
|
Answer» (a) Current through arm ABC, I = \(\frac{48}{10+20}\) = 1.6A Potential difference across B and C, = VB - VC = 1.6 × 20 = 32V As, Vc = 0 So, VB = +32V (b) Current through arm ADC, I' = \(\frac{48}{5+11}\) = 3A Potential difference across D and C, = VD - VC = 3 × 11 = 33V As, Vc = 0 So, VD = 33V (c) As VD > Vc, So direction of current will be from D to B. (d) The bridge will be balanced when : \(\frac{10}{R}\) = \(\frac{5}{11}\) or R = \(\frac{10\times 11}{5}\) = 22Ω. |
|