Question

A kΩ resistor is connected to an AC voltage source with an rms voltage of V. (a) What is the maximum potential difference across the resistor (in V)? V (b) What is the maximum current through the resistor (in A)? A (c) What is the rms current through the resistor (in A)? A (d) What is the average power dissipated by the resistor (in W)? W

Answer 1

Answer:

The answer is below

Explanation:

1.5 - kΩ resistor is connected to an AC voltage source with an  rms voltage of 120 V. (a) What is the maximum voltage across  the resistor? (b) What is the maximum current through the  resistor? (c) What is the rms current through the resistor?  d) What is the average power dissipated by the resistor?

Solution:

The rms value of current and voltage shows the alternating quantity of the voltage and current.

Given that V(rms) = 120 V, R = 1.5 kΩ

a) The maximum voltage across the resistor is given as:

$V_{max}=\sqrt{2}*V_{RMS}\\\\V_{max}=\sqrt{2}*120\\\\V_{max}=169. 7\ V$

b) The maximum current through the resistor is:

$I{max}=\frac{V_{max}}{R}\\\\ I{max}=\frac{169.7}{1.5*10^3}\\\\I{max}=0.113 \ ohm$

c) The rms current through the resistor is:

$I{rms}=\frac{V_{rms}}{R}\\\\ I{rms}=\frac{120}{1.5*10^3}\\\\I{rms}=0.08 \ ohm$

d) The average power dissipated by the resistor is:

$P_{av}=\frac{V_{rms}^2}{R}\\\\ P_{av}=\frac{120^2}{1500}\\\\P_{av}=9.6\ W$