Question

In the AC resistor circuit of Figure Q2.1, the voltage of the AC power source is

U = 100 sin(120pi x t) V .


what are

(a) the peak voltage of the AC power source

(b) the frequency of the alternating current in the circuit

(c) the instantaneous voltage of the power source at t = 20 ms

(d) the peak voltage across each resistor

(e) the instantaneous resistor voltages at t = 20 ms,

(f) to sketch the waveform of voltage in the first two periods

Answer 1

(a) 100 V

(b) 60 Hz

(c) 95.106 V

(d) 25 V, 75 V

(e) 23.776 V, 71.329 V

(f) see attached

Step-by-step explanation:

(a)

The peak value of the sine function is 1, so the peak value of U is ...

100·1 = 100 V.

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(b)

The argument of the sine function is 2πft, so f = 120πt/(2πt) = 60 Hz.

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(c)

When t=0.020, the voltage is ...

U = 100sin(120π·0.020) = 100 sin(2.4π) ≈ 95.106 V

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(d)

The resistors are in the ratio of 1 : 3, so divide the source voltage in that ratio.

The peak voltage across the 5Ω resistor is 1/4 of 100 V, or 25 V.

The peak voltage across the 15Ω resistor is 3/4 of 100 V, or 75 V.

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(e)

The same fractions as in (d) apply to the instantaneous voltage.

The voltage on the 5Ω resistor is 1/4·95.106 V ≈ 23.776 V

The voltage on the 15Ω resistor is 3/4·95.106 V ≈ 71.329 V

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(f)

See attached.