A battery of 9V is connected in series with resistors of 0.2omega ,0.3omega, 0.4omega, 0.5omega, and 12omega . How much current would flow through the 12omega resistor?

Asked by roysubarna2016 | 20th Feb, 2019, 01:30: PM

Expert Answer:

Since all resistor are connected in series so the current through all remain same
Equivalent resistance = 0.2+0.3+0.4+0.5+12= 13.4Ω
Using ohms law
Current I = V/R = 9/13.4= 0.67A

Answered by Ankit K | 20th Feb, 2019, 03:19: PM