While sitting in a house reading a book, we notice that every time the air conditioner comes on, the lights momentarily dim. Let us investigate this phenomenon using the single-phase three-wire circuit shown in Fig. 9.36a and some typical current requirements for a 10,000-Btu/h air

Question

While sitting in a house reading a book, we notice that every time the air conditioner comes on, the lights momentarily dim. Let us investigate this phenomenon using the single-phase three-wire circuit shown in Fig. 9.36a and some typical current requirements for a 10,000-Btu/h air conditioner, assuming a line resistance of 0.5 Ω.

Accepted Answer

[ h i n t ]: Technique
1. Find the resistance of the light bulb.
2. Use a large current source to represent the transient current of the ai conditioner and a small current source to represent the steady-state current.
3. Find the voltage drop across the light bulb during both the transient and steady-state operations.

The 60-W light bulb can be roughly modeled by its equivalent resistance:

[latex]P_{bulb} = \frac{V^{2}_{an}}{R_{bulb}}[/latex]

or

[latex]R_{bulb}[/latex] = 240 Ω

When the air conditioner unit first turns on, the current requirement is 40 A, as shown in Fig. 9.36b. As the compressor motor comes up to speed, the current requirement drops quickly to a steady-state value of 10 A, as shown in Fig. 9.36c. We will compare the voltage across the light fixture, [latex]V_{AN}[/latex], both at turn-on and in steady state.
Using superposition, let us first find that portion of [latex]V_{AN}[/latex] caused by the voltage sources. The appropriate circuit is shown in Fig. 9.36d. Using voltage division, we find that

[latex]V_{AN1}= V_{AN} \Big( \frac{R_{bulb}}{R_{bulb} + 2R_{L}} \Big)[/latex]

or

[latex]V_{AN1}[/latex] = 119.50 V rms

Figure 9.36e will yield the contribution to [latex]V_{AN}[/latex] caused by the 10-A steady-state current.
Using current division to calculate the current through the light bulb, we find that

[latex]V_{AN2} = - \{ I_{AB} \Big( \frac{R_{L}}{R_{bulb} + 2R_{L}} \Big) \} R_{bulb}[/latex]

or

[latex]V_{AN2}[/latex] = -4.98 V rms

Therefore, the steady-state value of [latex]V_{AN}[/latex] is

[latex]V_{AN} = V_{AN1} + V_{AN2}[/latex] = 114.52 V rms

At start-up, our expression for [latex]V_{AN2}[/latex] can be used with [latex]I_{AB}[/latex] = 40 A, which yields [latex]V_{AN2}[/latex] = -19.92 V rms. The resulting value for [latex]V_{AN}[/latex] is

[latex]V_{AN} = V_{AN1} + V_{AN2}[/latex] = 119.50 - 19.92 = 99.58 V rms

The voltage delivered to the light fixture at startup is 13% lower than the steady-state value, resulting in a momentary dimming of the lights.

Question 9.25: While sitting in a house reading a book, we notice that ever...

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