A Level Physics 9702 · A Level · Topic 21 · ~3 min read
Alternating currents
Describing sinusoidal alternating currents, peak and root-mean-square values, mean power in a resistor, and rectification with one diode, a bridge of four diodes, and smoothing with a capacitor.
Period, frequency, peak valueI = I₀ sin ωtr.m.s. valuesMean power = ½I₀²RHalf-wave rectificationFull-wave bridge rectifierSmoothing capacitorRipple
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In 30 seconds
- Sinusoidal a.c.: I = I₀ sin ωt and V = V₀ sin ωt, with ω = 2πf.
- r.m.s. value = peak ÷ √2. It's the steady direct current that would dissipate the same mean power in a resistor.
- Mean power in a resistor = ½I₀²R = Irms²R (half the peak power).
- Rectification: one diode gives half-wave; a bridge of four diodes gives full-wave. A capacitor in parallel with the load smooths the output; a larger RC gives less ripple.
By the end you can
- Understand period, frequency, peak value and root-mean-square values of alternating current and voltage.
- Use I = I₀ sin ωt and V = V₀ sin ωt.
- Use Irms = I₀/√2 and Vrms = V₀/√2, and mean power = ½ × maximum power for a resistive load.
- Explain half-wave and full-wave rectification and the effect of a smoothing capacitor.
1Describing a.c.
Peak and r.m.s.
Root-mean-square value — the value of the steady direct current (or voltage) that would produce the same mean power in a resistor.
I = I₀ sin ωt ; Irms = I₀/√2 ; Vrms = V₀/√2 DATA & FORMULAE
ω = 2πf- Mains in India: 230 V r.m.s. at 50 Hz, so the peak voltage is 230√2 = 325 V and the period 0.020 s.
Worked example
A supply V = 12 sin(100πt) is connected across a 6.0 Ω resistor. Find the frequency, the r.m.s. current and the mean power.
- ω = 100π ⇒ f = 50 Hz A1
- I₀ = 12/6.0 = 2.0 A ⇒ Irms = 1.4 A A1
- mean power = ½I₀²R = ½ × 4.0 × 6.0 = 12 W A1
Common mistakeUsing V₀²/R for mean power. That gives the peak power, twice the mean.
2Rectification and smoothing
Diodes and capacitors
- Half-wave: a single diode conducts only in one direction, so the negative half-cycles are lost.
- Full-wave: a bridge of four diodes routes current through the load in the same direction during both half-cycles.
- Smoothing: the capacitor charges near each peak and discharges through the load in between. Larger C or larger load R (larger time constant RC) means smaller ripple.
Common mistakeDrawing the smoothed output dropping to zero between peaks, or getting the bridge diode directions wrong. Practise drawing the bridge: both diodes on the output + side point towards it.
Linking questionCapacitor discharge is in Capacitance; diodes are introduced in Electricity; and generating a.c. by induction is in Magnetic fields.
Test yourself — cover the answers first
Quick check
- Find the r.m.s. value of a voltage with peak 20 V.
- How many diodes in a bridge rectifier?
- What does a smoothing capacitor do?
- Mean power is what fraction of peak power in a resistor?
- Find the period of 50 Hz a.c.
Exam-style
- An a.c. supply delivers 60 W (mean) to a 15 Ω resistor. Find the peak current.
- Explain why increasing the load resistance reduces the ripple.
- Write the equation for a 50 Hz current with r.m.s. value 3.0 A.
Show answers
- 14.1 V
- four
- reduces the ripple of the rectified output
- one half
- 0.020 s
- I_rms = 2.0 A ⇒ I₀ = 2.83 A
- the time constant RC increases, so the capacitor discharges more slowly between peaks
- I = 4.24 sin(100πt)
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