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Alternating currents questions
Alternating only means reversing, and the generator's version of it is a sine wave, so ac needs its own vocabulary. Peak, peak-to-peak, and the rms value, which is the steady dc equivalent for power in the same resistor. The oscilloscope is the instrument that lays a waveform out to be measured.
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.
Mark scheme
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.
Mark scheme
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.
Mark scheme
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 = 325.3 V (1)A sinusoidal alternating current has a peak value of 4.0 A. Calculate its rms value.
Mark scheme
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.
Mark scheme
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.
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.
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.
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.
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.
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.
Explain why rms values, rather than peak values, are normally quoted for alternating voltages and currents.
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.
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.
The heating effect of the 50 Hz mains supply reaches a maximum 100 times every second. Explain why.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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