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

Question

Halle Shelton

Physics

22 January, 15:15

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

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Answers (1)

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Ariella Dodson · Answer 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