SME
Circuit Analysis II · Problem type
RMS voltage from the peak
Use this for “A sine wave peaks at 170 V. Find its rms value and the average power it delivers to a 60 Ω resistor.”
Chart
Level Vᵣₘₛ = 120.2 meets the curve at ωt = 0.7854, ωt = 2.356
This is an example. Type your own values to start a new problem.
Vᵣₘₛ (V)
=
Vₘ (V)
2
P (W)
=
(
Vᵣₘₛ (V)
)
2
R (Ω)
Formulas
Vᵣₘₛ
=
Vₘ
÷
√
2
120.2082
=
170
÷
√
2
P
=
Vᵣₘₛ
2
R
240.8333
=
120.2082
2
60
Assumptions
•
The voltage is a sine wave, v = Vₘ sin ωt; a square wave’s rms equals its peak instead.
•
The rms value is the DC voltage that would heat the resistor at the same average rate.
•
The load is a resistor, so the average power is P = Vᵣₘₛ² ÷ R.
Step-by-step
WE KNOW
Vₘ = 170 V, R = 60 Ω
FIND
RMS voltage (Vᵣₘₛ), average power (P)
Step 1 · Find RMS voltage (Vᵣₘₛ)
Use Vᵣₘₛ = Vₘ ÷ √2
For
a
sine
wave
the
average
of
sin²
over
a
cycle
is
1
2
,
so
the
rms
is
the
peak
over
√
2
.
Vᵣₘₛ
=
170
÷
√
2
Vᵣₘₛ
=
170
1.414
Vᵣₘₛ = 120.20815 V
Step 2 · Find average power (P)
Use P = Vᵣₘₛ² ÷ R
The
rms
voltage
works
in
the
DC
power
rule,
P
=
V
2
R
,
to
give
the
average
power.
P
=
120.20815
2
60
P
=
14,450
60
P
=
240.8333
W
Check
Put the numbers back in. Both sides should match.
120.20815
=
170
÷
√
2
✓
240.8333
=
120.20815
2
60
✓
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