Question

In the given network, calculate:

1. effective resistance between points A and M, and
2. power supplied by the battery.

Answer 1

i.

So, `1/R_(eq) = 1/20 + 1/30`

= `(3 + 2)/60 = 5/60`

= `1/12`

⇒ R_eq = 12 Ω    ...[across A and M]

ii. Now, the equivalent resistance

`R_(eq) = 18 + (30 xx 20)/50 = 30  Omega`

So, `I = 6/30 = 1/5 A`

⇒ P = VI

= `6 xx 1/5`

= `6/5 W`

= 1.2 W