Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. what is the ratio of the power drawn from the battery for the two cases (i. e. p series: p parallel)? note: the battery supplies voltage v in both cases. a) 1: n b) 1: n2 c) n: 1 d) n2: 1

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R     (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2