Nitha Mulungu
Oct 14
A peak voltage of 250V AC is applied across a 30 ohms resistor. (1) Find the RMS current thought the resistor,(2) average power dissipated in the resistor,(3) amount of electrical energy use 2 min ?
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Oct 14
Given:
Peak voltage (Vp) = 250V
Resistance (R) = 30 ohms
Time (t) = 2 min = 120 seconds
(1) RMS current through the resistor:
The RMS current (Irms) can be calculated using the formula:
Irms = Vp / √2R
Irms = 250 / (√2 * 30)
Irms = 250 / 42.43
Irms ≈ 5.89 A
(2) Average power dissipated in the resistor:
The average power (P) dissipated in the resistor can be calculated using the formula:
P = Irms^2 * R
P = (5.89)^2 * 30
P ≈ 104.5 W
(3) Amount of electrical energy used in 2 min:
The total energy (E) used can be calculated using the formula:
E = P * t
E = 104.5 * 120
E = 12540 J
Therefore, the RMS current through the resistor is approximately 5.89 A, the average power dissipated in the resistor is approximately 104.5 W, and the amount of electrical energy used in 2 min is 12540 J.
Peak voltage (Vp) = 250V
Resistance (R) = 30 ohms
Time (t) = 2 min = 120 seconds
(1) RMS current through the resistor:
The RMS current (Irms) can be calculated using the formula:
Irms = Vp / √2R
Irms = 250 / (√2 * 30)
Irms = 250 / 42.43
Irms ≈ 5.89 A
(2) Average power dissipated in the resistor:
The average power (P) dissipated in the resistor can be calculated using the formula:
P = Irms^2 * R
P = (5.89)^2 * 30
P ≈ 104.5 W
(3) Amount of electrical energy used in 2 min:
The total energy (E) used can be calculated using the formula:
E = P * t
E = 104.5 * 120
E = 12540 J
Therefore, the RMS current through the resistor is approximately 5.89 A, the average power dissipated in the resistor is approximately 104.5 W, and the amount of electrical energy used in 2 min is 12540 J.
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