Saved Bookmarks
| 1. |
The galvanometer shown in the figure has resistance 10Omega. It is shunted by a series combination of a resistance S = 1Omega and an ideal cell of emf 2 V. A current 2 A passes as shown. |
|
Answer» The READING of the galvanometer is 1A. Let the currents be as shown in the figure. Applying KVL along ABCDA, we get `-10i-2 + (2-i) 1 = 0` `:. i = 0 ` Potential difference across = `(2-i)1 = 2XX1 = 2V` .
|
|