Circuit Analysis carries 11–17 items, tied with Mathematics for the widest range on the E&C specification. It is steady-state work: DC networks, and sinusoidal AC solved with phasors. Transients and resonance belong to Linear Systems.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Electrical and Computer
items · Circuit Analysis (DC and AC Steady State)11–17
problems on this page18
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What the area covers
Seven subtopics in the July 2020 specification, in plain terms:
KCL and KVL: currents summed at a node, voltages around a loop, with one sign convention throughout.
Series and parallel equivalents: reducing resistor networks, and the same reduction with complex impedances.
Thevenin and Norton: a network collapsed to one source and one impedance at a pair of terminals.
Node and loop analysis: writing the equations when reduction is not enough.
Waveforms: RMS and average values, frequency, phase and wavelength of periodic signals.
Phasors: sinusoids as complex numbers in polar or rectangular form.
Impedance: resistance and reactance combined as one complex number.
18 circuit analysis problems
0 / 18 answered · 0 correct
Question 1 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A sinusoidal voltage has a peak amplitude of 170 V. What is its RMS value?
Given
Peak voltage Vm = 170 V
Find
RMS voltage Vrms (V)
Solution
Vrms = Vm/√2
Vrms = 170 V/1.414 = 120.2 V
Vrms = Vm/√2 = 170/1.414 = 120.2 V.
This is the peak amplitude itself; the rms value of a sine is always smaller than its peak.
This multiplies the peak by √2 instead of dividing, as if converting rms to peak in the wrong direction.
This is the peak-to-peak value 2Vm, the largest of the amplitude measures.
Question 2 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A capacitor with capacitance 10 µF operates at 60 Hz. What is the magnitude of its capacitive reactance?
Given
Capacitance C = 10 µF
Frequency f = 60 Hz
Find
Capacitive reactance XC (Ω)
Solution
XC = 1/(ωC) = 1/(2πfC)
XC = 1/(2π × 60 Hz × 10 × 10−6 F) = 265.3 Ω
This is 1/(fC) with no 2π: hertz used in place of rad/s.
This is 1/(πfC): the 2 in 2πf was dropped, which doubles the reactance.
This is ω = 2πf = 377 rad/s reported as the reactance; the step 1/(ωC) was never taken.
This multiplies by √2, as if the 100 V rms were a peak value to be converted; rms values give S directly.
Question 7 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A load consumes 2400 W at 0.6 power factor lagging from a 120 V rms source. What is the reactive power consumed?
Given
Real power P = 2400 W
Power factor pf = 0.6 lagging
Source voltage V = 120 V rms
Find
Reactive power Q (VAR)
Solution
S = P/pf = 2400 W/0.6 = 4000 VA
Q = √(S2 − P2) = √(40002 − 24002) = 3200 VAR
This is P × pf = 2400 × 0.6, which is not a side of the power triangle.
This is S × (1 − pf) = 4000 × 0.4; reactive power follows Q = S sin θ, not a linear share of S.
This repeats the real power; P and Q are equal only at a power factor of 0.707.
S = P/pf = 2400/0.6 = 4000 VA, then Q = √(S2 − P2) = √(40002 − 24002) = 3200 VAR.
Question 8 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A full-wave rectified sinusoidal voltage has a peak value of 24 V. What is its average (DC) value?
Given
Peak voltage Vm = 24 V
Find
Average (DC) value Vavg (V)
Solution
Vavg = 2Vm/π
Vavg = 2 × 24 V/π = 15.28 V
This is Vm/π, the average of a half-wave rectified sine; full-wave rectification doubles it.
Vavg = 2Vm/π = 2 × 24/π = 15.28 V.
This is Vm/√2, the rms value of the waveform, not its average.
This is the peak; the average of any rectified sine sits well below the peak.
Question 9 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A load impedance ZL is connected to a source with Thevenin equivalent impedance Zth = 8 + j6 Ω. If ZL can be any complex impedance, what value gives maximum power transfer?
Solution
ZL = Zth* = 8 − j6 Ω
The reactances cancel, leaving RL = Rth (maximum power transfer theorem).
This conjugates the impedance but also swaps the real and imaginary parts; the resistance must stay 8 Ω.
Maximum power transfer needs ZL = Zth* (the complex conjugate): the reactances cancel and RL = Rth = 8 Ω.
This copies Zth; a load equal to the source impedance doubles the reactance in the loop instead of canceling it.
RL = |Zth| = 10 Ω is the best choice only when the load is restricted to a pure resistance; with a free complex load the conjugate delivers more.
Question 10 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A sinusoidal voltage has peak amplitude Vm = 170 V (which corresponds to 120 V rms). What is the rms value of the half-wave rectified version of this sinusoid?
Given
Peak voltage Vm = 170 V
Find
RMS value of the half-wave rectified wave (V)
Solution
Vrms = √((1/T)∫ v2 dt) = Vm/2 for a half-wave rectified sine
Vrms = 170 V/2 = 85.0 V
This is Vm/π, the average (DC) value of the half-wave rectified signal, not its rms value.
Half of each cycle is zero, so the mean square halves: Vrms = Vm/2 = 170/2 = 85.0 V.
This is the rms of the full, unrectified sine; removing half of each cycle lowers it.
This is the peak, which the rectified wave still reaches, but rms averages over the whole period.
Question 11 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
In a circuit with three series resistors of 10 Ω, 20 Ω, and 30 Ω connected to a 120 V DC source, what is the current flowing through the circuit?
Given
Resistance R1 = 10 Ω
Resistance R2 = 20 Ω
Resistance R3 = 30 Ω
Source voltage V = 120 V
Find
Circuit current I (A)
Solution
Rtotal = 10 + 20 + 30 = 60 Ω
I = V/Rtotal = 120 V/60 Ω = 2.0 A
This treats the resistors as parallel (Req = 5.45 Ω); a series string has the sum of the resistances.
This is 120/20, the current one 20 Ω resistor would draw alone; in series all three resistors limit the same current.
Series resistances add: R = 10 + 20 + 30 = 60 Ω, and I = V/R = 120/60 = 2.0 A.
This is R/V = 60/120, Ohm's law inverted.
Question 12 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
At a node in a circuit, three currents enter with magnitudes 5 A, 3 A, and 2 A, and two currents leave. If one leaving current is 7 A, what is the magnitude of the other leaving current according to KCL?
Given
Entering current I1 = 5 A
Entering current I2 = 3 A
Entering current I3 = 2 A
Leaving current I4 = 7 A
Find
Other leaving current I5 (A)
Solution
∑Iin = ∑Iout
5 + 3 + 2 = 7 + I5 → I5 = 3 A
KCL: total in = 5 + 3 + 2 = 10 A, so the second leaving current is 10 − 7 = 3 A.
This splits the 10 A of entering current equally between the two leaving branches and ignores the given 7 A.
This is the total entering current; one of the leaving branches already carries 7 A of it.
This adds the known leaving current to the entering total instead of subtracting it.
Question 13 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
In a closed loop containing a 12 V battery, a 3 Ω resistor, and a 1 Ω resistor in series, what is the voltage drop across the 3 Ω resistor?
Given
Battery voltage V = 12 V
Resistance R1 = 3 Ω
Resistance R2 = 1 Ω
Find
Voltage across the 3 Ω resistor (V)
Solution
I = V/(R1 + R2) = 12 V/4 Ω = 3 A
V3Ω = I·R1 = 3 A × 3 Ω = 9 V
This is the drop across the 1 Ω resistor, I × 1 Ω = 3 A × 1 Ω.
This splits 12 V equally between the resistors, which holds only for equal resistances.
I = 12/(3 + 1) = 3 A, and V = I × 3 Ω = 9 V; the two drops (9 V and 3 V) sum to 12 V as KVL requires.
This is the whole battery voltage; part of it drops across the 1 Ω resistor.
Question 14 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
An AC circuit has a voltage of 120 V rms and draws a current of 10 A rms with a power factor of 0.8 lagging. What is the real power consumed?
Given
Voltage Vrms = 120 V
Current Irms = 10 A
Power factor pf = 0.8 lagging
Find
Real power P (W)
Solution
P = Vrms · Irms · cos θ
P = 120 V × 10 A × 0.8 = 960 W
This is the reactive power V·I·sin θ = 1200 × 0.6, in var, not watts.
P = Vrms·Irms·pf = 120 × 10 × 0.8 = 960 W.
This is the apparent power V·I in VA; the power factor was not applied.
This divides by the power factor instead of multiplying.
Question 15 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
Two impedances Z1 = 4 + j3 Ω and Z2 = 2 − j1 Ω are connected in series. What is the total impedance?
Given
Impedance Z1 = 4 + j3 Ω
Impedance Z2 = 2 − j1 Ω
Find
Total series impedance Z (Ω)
Solution
Z = Z1 + Z2 = (4 + 2) + j(3 − 1) = 6 + j2 Ω
Series impedances add part by part: (4 + 2) + j(3 − 1) = 6 + j2 Ω.
This is Z1Z2/(Z1 + Z2), the parallel combination; series impedances simply add.
This adds the reactances as magnitudes (3 + 1) and ignores the minus sign of the capacitive −j1.
This is Z1 − Z2: both parts were subtracted.
Question 16 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A circuit consists of a 120 V rms voltage source in series with a 30 Ω resistor and a capacitor with reactance 40 Ω. What is the power factor of the circuit?
Given
Source voltage V = 120 V rms
Resistance R = 30 Ω
Capacitive reactance XC = 40 Ω
Find
Power factor
Solution
|Z| = √(R2 + XC2) = √(302 + 402) = 50 Ω
pf = cos θ = R/|Z| = 30/50 = 0.60 leading
This is 30/(30 + 40): R and X were added arithmetically instead of as legs of the impedance triangle.
|Z| = √(302 + 402) = 50 Ω and pf = R/|Z| = 30/50 = 0.60 (leading, since the load is capacitive).
This is R/X = 30/40, which is tan of the complementary angle, not cos θ.
This is X/|Z| = 40/50 = sin θ, the reactive factor, not the power factor.
Question 17 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A circuit consists of an ideal voltage source in series with a 20 Ω resistor and a 60 Ω resistor. What is the Thevenin equivalent resistance seen from the terminals of the 60 Ω resistor after that resistor is removed?
Given
Series resistance R1 = 20 Ω
Load resistance R2 = 60 Ω (removed)
Find
Thevenin resistance Rth (Ω)
Solution
Rth: voltage source → short circuit, load removed
Rth = 20 Ω
This keeps the load and treats the source as open: 20 + 60 = 80 Ω; an ideal voltage source is replaced by a short.
This is the removed load itself, which is not part of the Thevenin equivalent.
Remove the 60 Ω load and short the ideal source: the only resistance seen from the open terminals is the 20 Ω resistor.
This is 20 ∥ 60: the 60 Ω load was left in place, but Rth is found with the load removed.
Question 18 of 18 · Circuit Analysis (DC and AC Steady State) · 11–17 items
In a nodal analysis, node 2 is held at 10 V by a source and the reference node is ground. The KCL equation at node 1 is V1/5 + (V1 − 10)/10 = 2. What is V1?
Given
Node equation V1/5 + (V1 − 10)/10 = 2
Find
Node voltage V1 (V)
Solution
V1/5 + (V1 − 10)/10 = 2
2V1 + V1 − 10 = 20 → V1 = 30/3 = 10.0 V
This puts the 2 A on the wrong side of the KCL equation (as a current leaving the node), which flips the sign.
This writes the branch term as (V1 + 10)/10, a sign error on the source voltage.
This clears the fractions by multiplying by 10 on the left side only: 2V1 + V1 − 10 = 2.
Multiply both sides by 10: 2V1 + V1 − 10 = 20, so 3V1 = 30 and V1 = 10.0 V.
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How AC problems are solved
AC steady-state problems reduce to complex arithmetic. Turn each source into a phasor, replace each element by its impedance, solve the network as if it were DC, and convert back to a magnitude and angle.
ZR = R ZL = jωL ZC = 1/(jωC) = −j/(ωC)
S = P + jQ, |S|2 = P2 + Q2
The calculator does the heavy lifting. A model whose complex mode you know cold saves time on every phasor item; practice on the exact approved model you will bring (calculator rules). Formulas are in the Electrical and Computer Engineering chapter of the FE Reference Handbook.
Common traps in circuit analysis
RMS or peak. Power and meter readings are RMS values; a sine’s RMS value is Vm/√2. Check which one the stem gives before you multiply.
The capacitor’s sign. Its impedance is −j/(ωC). Writing −jωC makes the reactance grow with frequency, which is the inductor’s behavior.
ω or f. Reactance takes ω = 2πf in rad/s; plugging in hertz is off by a factor of about 6.28.
Degrees or radians. Phasor angles are given in degrees; a calculator left in radian mode returns a wrong angle with no warning.
Passive sign convention. Power is absorbed when current enters the + terminal; a source that delivers power shows a negative product.
Thevenin with a dependent source. Do not switch it off. Keep it, apply a test source, and take Vtest/Itest.
Sources: the sinusoidal steady-state and power chapters of a circuits text such as Nilsson and Riedel, Electric Circuits.
How it connects to other areas
Circuit skills carry four other E&C areas: diode and op-amp circuits in Electronics, RLC transients and filters in Linear Systems, three-phase circuits and transformer equivalents in Power Systems, and transmission lines in Communications. AC power ends here and continues in Power Systems.
This page is also where other disciplines in the hub send electricity: the Electricity and Magnetism area of FE Mechanical (5–8 items there) links here. Complex-number algebra itself is part of Mathematics, practiced on the FE practice test.