Saved Bookmarks
| 1. |
Find the value of R in fig. so that there is no current in the 15 Omega resistor |
|
Answer» Solution :This is the WHEATSTONE bridge with the galvanometer replaced by `15 OMEGA` RESISTOR. The bridge is balanced because there is no CURRENT is 15 `Omega` resistor, hence, `20//10 = 40//R` `THEREFORE R= (40 xx 10 Omega)/(20) = 20 Omega` |
|