A 150 ohm resistor and inductor of inductance L are connected in series to a 200 V, 50 Hz source of negligilbe impedance. The current comes to 1 A. when the source is changed to 400 V, 100 Hz, the current will be

Options

A. less than 1.0 A
B. 1.0 A
C. between 1 A and 2 A
D. between 4A and 2 A

Answer

C

Explanation

Here, `R = 150 Omega, V = 200 V, v = 50 HZ`,
`I = 1A, V' = 400 V. v' = 100 Hz, I' = ?`
`I = (200)/(sqrt(R^(2) + X_(L)^(2))) = 1, sqrt(R^(2) + X_(L)^(2)) = 200`
`(150)^(2) + X_(L)^(2) = (2000)^(2)`
`X_(L)^(2) = 17,500`
`X_(L) = 2 pi v L, (X'_(L))/(X_(L)) = (2 pi v' L)/(2 pi v L) = 0`, so `X'_(L) = 2 X_(L)`
`X'_(L) = 4 X_(L)^(2) = 4 xx 175000 = 70,000`
`Z' = sqrt(R^(2) + X_(L)^(2)) = sqrt((150)^(2) + 70,000) = 304 Omega`
`I' = (400 V)/(304 Omega) = 1.3 A` (between 1 A and 2 A)