In an experiment to measure the active power drawn by a single-phase RL Load connected to an AC source through a \(2\,\Omega\) resistor, three voltmeters are connected as shown in the figure below. The voltmeter readings are as follows: \( V_{{Source}} = 200\,{V}, \quad V_R = 9\,{V}, \quad V_{{Load}} = 199\,{V}. \) Assuming perfect resistors and ideal voltmeters, the Load-active power measured in this experiment, in W, is __________ (round off to one decimal place).
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In AC circuits, when only voltmeter readings are available, use the vector relationship between voltages to calculate the power factor. Then use \( P = VI\cos\phi \) to estimate active power consumed by the load.
We are given:
\[
V_{{Source}} = 200~{V}, \quad V_R = 9~{V}, \quad V_{{Load}} = 199~{V}, \quad R = 2~\Omega
\]
Step 1: Find Current using voltage across the resistor:
\[
V_R = I \cdot R \Rightarrow I = \frac{V_R}{R} = \frac{9}{2} = 4.5~{A}
\]
Step 2: Use current and load voltage to compute active power:
\[
P = V_{{Load}} \cdot I \cdot \cos\theta
\]
But since we don’t have power factor \( \cos\theta \), and we're being asked for measured active power, and only voltmeters are used (no wattmeter), the standard method used in such experiments is:
\[
P = V_R \cdot I = I^2 \cdot R
\Rightarrow P = (4.5)^2 \cdot 2 = 20.25 \cdot 2 = \boxed{40.5~{W}}
\]
Wait! That’s power dissipated in the 2 ohm resistor, not in the load.
Let’s correct:
Load voltage is \( V_{{Load}} = 199~{V} \)
Current through load is same: \( I = 4.5~{A} \)
So power consumed by the load:
\[
P_{{Load}} = V_{{Load}} \cdot I \cdot \cos\theta
\]
But we don't know \( \cos\theta \). So how is power being estimated?
Actually, this is a known two-voltmeter method to compute power factor and power:
- From:
\[
V_{{Source}}^2 = V_R^2 + V_{{Load}}^2 + 2 V_R V_{{Load}} \cos\phi
\]
Substitute:
\[
200^2 = 9^2 + 199^2 + 2 \cdot 9 \cdot 199 \cdot \cos\phi
\]
\[
40000 = 81 + 39601 + 3582 \cos\phi
\]
\[
40000 = 39682 + 3582 \cos\phi
\]
\[
\cos\phi = \frac{40000 - 39682}{3582}
\]
\[
\cos\phi = \frac{318}{3582} \approx 0.0888
\]
Now compute:
\[
P_{{Load}} = V_{{Load}} \cdot I \cdot \cos\phi = 199 \cdot 4.5 \cdot 0.0888 \approx 79.6~{W}
\]
\[
\boxed{P_{{Load}} \approx 79.6~{W}}
\]
