The peak value of the ac current in a circuit is 6.0 A, and peak value of the ac voltage applied to the circuit is 8.9 V. What is the average power P delivered to the circuit?
The average power dissipated in a 47-Ù resistor is 5.9 W. What is the peak value I0 of the ac current in the resistor?
Average power is half the product of peak voltage and peak current.
For your second question
Avg. Power = (1/2)Io^2*R
Solve for Io
Peak voltages and currents are usually not used in AC calculations; instead, r.m.s. (root mean square) values are used. They are 0.707 (1/sqrt2) times the peak value
To find the average power delivered to the circuit, you need to use the formula:
P_avg = (V_peak * I_peak) / √2
where:
P_avg is the average power
V_peak is the peak value of the voltage
I_peak is the peak value of the current
In this case, V_peak = 8.9 V and I_peak = 6.0 A. Now, let's calculate the average power:
P_avg = (8.9 V * 6.0 A) / √2
To simplify, let's divide the value by √2:
P_avg = (8.9 V * 6.0 A) / 1.414
P_avg ≈ 37.70 V * A / 1.414
Using a calculator, we can calculate the result:
P_avg ≈ 26.70 W
Therefore, the average power delivered to the circuit is approximately 26.70 Watts (W).
To calculate the average power delivered to the circuit, you can use the formula:
P = (Vp * Ip) / 2
where P is the average power, Vp is the peak voltage, and Ip is the peak current.
Given that the peak value of the ac current is 6.0 A (Ip = 6.0 A) and the peak value of the ac voltage applied to the circuit is 8.9 V (Vp = 8.9 V), you can substitute the values into the formula:
P = (8.9 V * 6.0 A) / 2
P = 53.4 VA / 2
P = 26.7 W
Therefore, the average power delivered to the circuit is 26.7 watts.