The average power delivered to an impedance (4 – j3) Ω by a current 5 cos (100πt + 100) A is

Average Power Calculation for Impedance

To determine the average power delivered to an impedance by an alternating current, we need to understand the relationship between the resistive part of the impedance and the root-mean-square (RMS) value of the current. The average power in an AC circuit is dissipated only by the resistive component of the impedance, while the reactive component (inductance or capacitance) does not consume average power.

Impedance and Current Parameters

The problem provides us with the following parameters:

- The impedance, denoted as $Z$, is given in rectangular form: $Z=(4-j3)\Omega$.
- The current, denoted as $i(t)$, is given as a sinusoidal function of time: $i(t)=5\cos (100\pi t+100)\text{A}$.

From the impedance $Z=(4-j3)\Omega$, we can identify the resistive part and the reactive part:

- The resistive component (resistance) $R=4\Omega$.
- The reactive component (reactance) $X=-3\Omega$ (indicating a capacitive reactance).

From the current expression $i(t)=5\cos (100\pi t+100)\text{A}$, we can identify the peak amplitude of the current:

- The peak current $I_m=5\text{A}$.

Current RMS Value Determination

The average power dissipated in a resistor is calculated using the RMS value of the current. The RMS value of a sinusoidal current is related to its peak value by the formula:

$$I_{rms}=\frac{I_m}{\sqrt{2}}$$

Substituting the peak current $I_m=5\text{A}$ into the formula:

$$I_{rms}=\frac{5}{\sqrt{2}}\text{A}$$

Power Delivered Calculation

The average power $P_{avg}$ delivered to the impedance is exclusively dissipated by its resistive component $R$. The formula for average power using RMS current and resistance is:

$$P_{avg}=I_{rms}^{2}R$$

Now, we substitute the calculated RMS current and the identified resistance into this formula:

$$P_{avg}={\left(\frac{5}{\sqrt{2}}\right)}^2\times 4$$

First, calculate the square of the RMS current:

$${\left(\frac{5}{\sqrt{2}}\right)}^2=\frac{5^2}{(\sqrt{2}{)}^2}=\frac{25}{2}$$

Now, multiply this by the resistance $R=4\Omega$:

$$P_{avg}=\frac{25}{2}\times 4$$

$$P_{avg}=25\times 2$$

$$P_{avg}=50\text{W}$$

Therefore, the average power delivered to the impedance is 50 W.