Question

A resistor is connected across an oscillating emf. The peak current through the resistor is 2.0 A. What is the peak current if:

a. The resistance R is doubled?
b. The peak emf εo is doubled?
c. The frequency ω is doubled?

Answer 1

(a) When the resistance R is doubled, I = 1 A

(b) When the peak emf εo is doubled, I = 4 A

(c) When the frequency ω is doubled, I = 2 A

Step-by-step explanation:

Given;

peak current through the resistor, I = 2.0 A

According to ohms law the peak current through the circuit is given by;

`I = (V)/(R)`

(a) When the resistance R is doubled;

`I = (V_R)/(R) \\\\I_1R_1 = I_2R_2\\\\I_2 = (I_1R_1)/(R_2) \\\\I_2 = (2*R_1)/(2R_1) \\\\I_2 = 1 \ A`

(b)When the peak emf εo is doubled

`I = (V)/(R) = (\epsilon_o)/(R) \\\\R = (\epsilon_ o)/(I) \\\\(\epsilon_ o_1)/(I_1) = (\epsilon_ o_2)/(I_2) \\\\I_2 = (\epsilon_ o_2 *I_1)/(\epsilon _o_1) \\\\I_2 = (2 \epsilon_ o_1 *2)/(\epsilon _o_1) \\\\I_2 = 4 \ A`

(c) When the frequency ω is doubled

Peak current through resistor is independent of frequency

I₂ = 2.0 A