Saved Bookmarks
| 1. |
Three resistances of 2Ω each are connected to a 2V battery as shown in the front figure. Then ........ |
|
Answer» Given circuit resistance are parallel combination then \(\frac1R=\frac1R_1-\frac1R_2+\frac1R_2\) \(\frac1R=\frac12+\frac12+\frac12\) \(\frac1R=\frac32\) R = 2/3 We use V = iR i = V/R i = \(\frac{2\times3}2\) = 3 i = 3 ampeare Then I1 = 2A it divided two branches then I1 = 1 + 1 Then I2 = 2A |
|