1 Answer
In an RL circuit (resistor and inductor in series), the apparent power (\( S \)) is given by:
\[
S = V \times I
\]
Where:
However, in an RL circuit, because the voltage and current are not in phase, the formula is often expressed in terms of the impedance (\( Z \)) and the phase angle (\( \phi \)) between the voltage and the current:
\[
S = V_{\text{rms}} \times I_{\text{rms}} = V_{\text{rms}} \times \frac{V_{\text{rms}}}{Z}
\]
This simplifies to:
\[
S = \frac{V_{\text{rms}}^2}{Z}
\]
Where:
So, apparent power in an RL circuit takes into account both the resistive and inductive components of the impedance.
\[
S = V \times I
\]
Where:
- \( V \) is the rms voltage across the circuit
- \( I \) is the rms current flowing through the circuit
However, in an RL circuit, because the voltage and current are not in phase, the formula is often expressed in terms of the impedance (\( Z \)) and the phase angle (\( \phi \)) between the voltage and the current:
\[
S = V_{\text{rms}} \times I_{\text{rms}} = V_{\text{rms}} \times \frac{V_{\text{rms}}}{Z}
\]
This simplifies to:
\[
S = \frac{V_{\text{rms}}^2}{Z}
\]
Where:
- \( Z = \sqrt{R^2 + X_L^2} \) is the total impedance of the RL circuit, with \( R \) being the resistance and \( X_L = \omega L \) being the inductive reactance.
- \( \omega \) is the angular frequency (\( \omega = 2 \pi f \), where \( f \) is the frequency of the AC source).
- \( L \) is the inductance of the coil.
So, apparent power in an RL circuit takes into account both the resistive and inductive components of the impedance.