Derivation of Active Power in an R-L Series Circuit
1. Circuit Quantities and Phasor Relationship
Consider an alternating sinusoidal voltage applied across a series combination of resistance (\(R\)) and inductance (\(L\)).
Let the instantaneous supply voltage be taken as the reference phasor:
$$v(t) = V_m \sin(\omega t)$$
In an inductive circuit, the circuit current lags behind the applied voltage by a phase angle \(\phi\) (where \(0 < \phi < 90^\circ\)). Therefore, the instantaneous current is:
$$i(t) = I_m \sin(\omega t - \phi)$$
where:
- \(V_m\) = Maximum (peak) value of voltage
- \(I_m\) = Maximum (peak) value of current
- \(\omega\) = Angular frequency in \(\text{rad/s}\)
- \(\phi = \tan^{-1}\left(\frac{X_L}{R}\right)\) = Phase angle of the circuit
2. Step-by-Step Derivation
Step 1: Instantaneous Power (\(p\))
The instantaneous power supplied to the circuit at any instant is the product of instantaneous voltage and instantaneous current:
$$p = v(t) \cdot i(t) = [V_m \sin(\omega t)] \cdot [I_m \sin(\omega t - \phi)]$$
$$p = V_m I_m \sin(\omega t) \sin(\omega t - \phi)$$
Step 2: Applying Trigonometric Identity
Using the identity \(2 \sin A \sin B = \cos(A - B) - \cos(A + B)\):
$$p = \frac{V_m I_m}{2} \Big[ \cos\big(\omega t - (\omega t - \phi)\big) - \cos\big(\omega t + (\omega t - \phi)\big) \Big]$$
$$p = \frac{V_m I_m}{2} \big[ \cos(\phi) - \cos(2\omega t - \phi) \big]$$
$$p = \frac{V_m I_m}{2} \cos(\phi) - \frac{V_m I_m}{2} \cos(2\omega t - \phi)$$
Step 3: Average (Active) Power over a Complete Cycle (\(P\))
The total average power (Active Power, \(P\)) taken by the circuit over one complete cycle (\(T = \frac{2\pi}{\omega}\)) is the time-average of the instantaneous power:
$$P = \frac{1}{2\pi} \int_{0}^{2\pi} p \, d(\omega t)$$
$$P = \frac{1}{2\pi} \int_{0}^{2\pi} \left[ \frac{V_m I_m}{2} \cos(\phi) - \frac{V_m I_m}{2} \cos(2\omega t - \phi) \right] d(\omega t)$$
Since the second term \(\cos(2\omega t - \phi)\) is a double-frequency sinusoidal component, its average value over a full period is zero:
$$\frac{1}{2\pi} \int_{0}^{2\pi} \cos(2\omega t - \phi) \, d(\omega t) = 0$$
Thus, only the constant first term remains:
$$P = \frac{V_m I_m}{2} \cos(\phi)$$
Step 4: Expressing in RMS Values
Writing \(\frac{V_m I_m}{2}\) in terms of RMS quantities (\(V = \frac{V_m}{\sqrt{2}}\) and \(I = \frac{I_m}{\sqrt{2}}\)):
$$P = \left(\frac{V_m}{\sqrt{2}}\right) \left(\frac{I_m}{\sqrt{2}}\right) \cos(\phi)$$
3. Alternate Expressions of Active Power
-
In Terms of Resistance (\(R\)):
From the impedance triangle, the power factor is \(\cos(\phi) = \frac{R}{Z}\), and the applied voltage is \(V = I \cdot Z\). Substituting these:
$$P = (I \cdot Z) \cdot I \cdot \left(\frac{R}{Z}\right) = I^2 R \quad \text{Watts}$$
(This proves that in an R-L series circuit, active power is consumed exclusively by the resistance; the pure inductor consumes zero average active power).
- Active Power (True Power): \(P = V I \cos\phi = I^2 R\) [Unit: Watts / kW]
- \(\cos\phi\): Power Factor of the circuit (\(\cos\phi = \frac{R}{Z}\), lagging)
- Inductor Power: Pure inductance consumes no net active power over a complete cycle.
Efficiency Calculation of a DC Shunt Motor by Swinburne’s Test
Problem Statement
A 200 V, 14.92 kW DC shunt motor when tested by the Swinburne’s method gave the following results:
Running Light (No-Load): Armature current = 6.5 A, Field current = 2.2 A.
Locked Armature Test: Current = 70 A when a potential difference of 3 V was applied across the brushes.
Objective: Estimate the efficiency of the motor when operating under full-load conditions.
Given Data
Terminal Voltage ($V$) = 200 V
Full-Load Shaft Output ($P_{\text{out}}$) = 14.92 kW = 14,920 W
No-load Armature Current ($I_{a0}$) = 6.5 A
Shunt Field Current ($I_{sh}$) = 2.2 A
Voltage applied during locked-rotor test ($V_{br}$) = 3 V
Armature current during locked-rotor test ($I_{br}$) = 70 A
Step 1: Determination of Armature Resistance ($R_a$)
From the locked armature test across the brushes:
Step 2: Determination of Constant Losses ($W_c$)
Under running light (no-load) conditions:
No-load input power to armature:
$$P_{a0} = V \times I_{a0} = 200 \times 6.5 = 1300\text{ W}$$No-load armature copper loss:
$$P_{cu0} = I_{a0}^2 \times R_a = (6.5)^2 \times 0.04286 = 1.81\text{ W}$$Iron, friction, and windage losses (Stray Mechanical Losses):
$$W_m = P_{a0} - P_{cu0} = 1300 - 1.81 = 1298.19\text{ W}$$Shunt field copper loss:
$$W_{sh} = V \times I_{sh} = 200 \times 2.2 = 440\text{ W}$$Total constant losses ($W_c$):
$$W_c = W_m + W_{sh} = 1298.19 + 440 = 1738.19\text{ W}$$
Step 3: Full-Load Analysis & Efficiency
Method 1: Standard Approximate Method (Common in Exam Evaluations)
Assuming initial motor input power is approximately equal to the rated mechanical output:
Approximate full-load line current ($I_{FL}$):
$$I_{FL} \approx \frac{P_{\text{out}}}{V} = \frac{14,920}{200} = 74.6\text{ A}$$Full-load armature current ($I_a$):
$$I_a = I_{FL} - I_{sh} = 74.6 - 2.2 = 72.4\text{ A}$$Full-load armature copper loss:
$$P_{cu,FL} = I_a^2 \times R_a = (72.4)^2 \times 0.04286 = 224.66\text{ W}$$Total full-load losses:
$$W_{\text{total}} = W_c + P_{cu,FL} = 1738.19 + 224.66 = 1962.85\text{ W}$$Full-load electrical input power:
$$P_{\text{in}} = P_{\text{out}} + W_{\text{total}} = 14,920 + 1962.85 = 16,882.85\text{ W}$$Full-load efficiency ($\eta$):
$$\eta = \left( \frac{P_{\text{out}}}{P_{\text{in}}} \right) \times 100 = \left( \frac{14,920}{16,882.85} \right) \times 100 \approx \mathbf{88.37\%}$$