Answer:
(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
Explanation:
Given;
peak current through the resistor, I = 2.0 A
According to ohms law the peak current through the circuit is given by;
[tex]I = \frac{V}{R}[/tex]
(a) When the resistance R is doubled;
[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]
(b)When the peak emf εo is doubled
[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]
(c) When the frequency ω is doubled
Peak current through resistor is independent of frequency
I₂ = 2.0 A
