Saved Bookmarks
| 1. |
The resistance in the four arms of a Wheatstone network in cyclic order are 5Omega, 2Omega, 6Omega and 15Omega. If a current of 2.8 A enters the junction of 5Omega and 15Omega, then the current through 2Omega resistor is |
|
Answer» 1.5 A Resistance of the upper arm = `5Omega+2Omega=7Omega` Resistance of the lower arm = `15Omega+6Omega=21Omega` Current through the upper arm `I_(1)=(2.8Axx21Omega)/(7Omega+21Omega)=2.1A` Hence, the current through the `2Omega` resistor is 2.1 A. |
|