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Questions on alternating currents
An alternating current reverses direction periodically, and for a sinusoid the rms value is the peak divided by root two, or 0.707 of it. That rms value is the steady dc delivering the same mean power in the same resistor, and mains is 230 V rms with a 325 V peak. On an oscilloscope, the y-gain and time base convert divisions to volts and seconds.
18 original questions · 49 marks · the alternating currents notes · Magnetic fields
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Define the root mean square (rms) value of an alternating current.
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The value of the steady direct current (1); that would dissipate the same average power in a given resistor as the alternating current (1).A sinusoidal alternating voltage has a peak value of 10 V. Calculate its rms value.
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Vrms = Vpeak/√2 = 10/√2 (1)
Vrms = 7.07 V (1)A sinusoidal alternating voltage has a peak value of 15 V. State its peak-to-peak value.
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Peak-to-peak value = 2 × 15 = 30 V (1).The frequency of the UK mains supply is 50 Hz. Calculate its period and its angular frequency.
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T = 1/f = 1/50 = 0.020 s (1)
ω = 2πf = 2π × 50 = 314 rad s−1 (1)On an oscilloscope with the time base set to 5.0 ms per division, one full cycle of a sinusoidal signal spans 4.0 divisions. Calculate the frequency of the signal.
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T = 4.0 × 5.0 ms = 20 ms (1)
f = 1/T = 1/0.020 = 50 Hz (1)The UK mains supply has an rms voltage of 230 V. Calculate the peak voltage of the supply.
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Vpeak = √2 × Vrms = √2 × 230 (1)
Vpeak = 325 V (1)A sinusoidal alternating current has a peak value of 4.0 A. Calculate its rms value.
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Irms = Ipeak/√2 = 4.0/√2 (1)
Irms = 2.83 A (1)A sinusoidal alternating supply of rms voltage 12 V is connected across a 6.0 Ω resistor. Calculate the rms current, the mean power dissipated and the peak power.
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Irms = Vrms/R = 12/6.0 = 2.0 A (1)
Mean power = VrmsIrms = 12 × 2.0 = 24 W (1)
Peak power = VpeakIpeak = 48 W (1)
Peak power is twice the mean power (1).An oscilloscope trace shows a sinusoidal alternating voltage with a peak value of 4.0 V. Calculate the rms voltage.
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Vrms = Vpeak/√2 = 4.0/√2 (1)
Vrms = 2.83 V (1)An alternating pd is described by v = 325 sin(100πt), with v in volts and t in seconds. State the peak voltage, and determine the frequency and the rms voltage of the supply.
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Peak voltage = 325 V (1)
ω = 100π rad s−1, so f = ω/2π = 50 Hz (1)
Vrms = 325/√2 = 230 V (1)The current in a 15 Ω resistor is sinusoidal with a peak value of 2.6 A. Calculate the mean power dissipated in the resistor.
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Irms = 2.6/√2 = 1.84 A (1)
P = Irms2R = 1.842 × 15 (1)
P = 51 W (1)On an oscilloscope screen, the trace of a sinusoidal voltage spans 6.4 divisions from its lowest point to its highest point. The y-gain is set to 0.50 V per division. Determine the rms voltage of the signal.
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Peak-to-peak voltage = 6.4 × 0.50 = 3.2 V (1)
Vpeak = 3.2/2 = 1.6 V (1)
Vrms = 1.6/√2 = 1.13 V (1)On an oscilloscope, one complete cycle of a sinusoidal signal spans 4.0 divisions with the time base set to 5.0 ms per division. The amplitude of the trace is 3.0 divisions with the voltage gain set to 2.0 V per division. Determine the period, the frequency and the rms voltage of the signal.
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T = 4.0 × 5.0 ms = 20 ms (1)
f = 1/T = 50 Hz (1)
Vpeak = 3.0 × 2.0 = 6.0 V (1)
Vrms = 6.0/√2 = 4.24 V (1)A 100 Ω resistor is connected to the 230 V rms mains supply. Calculate the mean power dissipated in the resistor and explain why a steady 230 V DC supply would dissipate the same power.
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Mean power = Vrms2/R = 2302/100 (1)
P = 529 W (1)
The rms value is defined as the DC value giving the same average power in a resistor, so 230 V DC gives the same power (1).Explain why rms values, rather than peak values, are normally quoted for alternating voltages and currents.
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The rms value is the steady (DC) equivalent for power dissipation (1); it represents the effective heating or working value of the supply over a cycle (1); so AC and DC can be compared directly and average power calculated simply as P = VrmsIrms (1).A camping inverter supplies 12 V rms and a maximum rms current of 7.0 A. A camper wants a heating element that delivers at least 60 W and considers two elements, of resistance 2.0 Ω and 3.0 Ω. Deduce which element, if either, the camper should choose.
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P = Vrms2/R: for 2.0 Ω, P = 122/2.0 = 72 W (1)
For 3.0 Ω, P = 122/3.0 = 48 W, below 60 W (1)
Current check for the 2.0 Ω element: Irms = 12/2.0 = 6.0 A, within the 7.0 A limit (1)
Only the 2.0 Ω element meets the 60 W requirement, so choose it (1)A lamp of constant resistance 6.0 Ω is connected first to a 12 V dc battery and then to a sinusoidal supply of peak voltage 12 V. Deduce which supply makes the lamp brighter and state the factor between the two powers.
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dc: P = V2/R = 122/6.0 = 24 W (1)
ac: Vrms = 12/√2 = 8.5 V (1)
P = Vrms2/R = 8.52/6.0 = 12 W (1)
The dc supply makes the lamp brighter, with twice the power, because the ac supply spends most of each cycle below its peak (1)The heating effect of the 50 Hz mains supply reaches a maximum 100 times every second. Explain why.
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The power dissipated is proportional to v2, which is positive in both half-cycles (1); the power is a maximum whenever the voltage reaches +peak or −peak, which happens twice per cycle (1); so the heating peaks at 2 × 50 = 100 times per second (1).
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