Circuit Analysis 1: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Series & parallel resistors

Resistors in series
Req = R1 + R2 + …
Resistors in parallel
1/Req = 1/R1 + 1/R2 + …
Two in parallel
Req = R1R2 / (R1 + R2)
Ohm's law
V = I·R
Power
P = V·I = I²R = V²/R

Method

  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. Repeat until one resistor is left between the terminals.

Nodal analysis

Kirchhoff’s current law
ΣIin = ΣIout
Branch current
I = (Va − Vb)/R
Reference node
Vground = 0

Method

  1. Choose a reference node (ground) and label the other node voltages.
  2. Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
  3. Express every branch current as (Va − Vb)/R.
  4. Solve the system of equations, then find any current you need.

Mesh analysis

Kirchhoff’s voltage law
ΣV = 0 around every loop
Shared resistor
current = I1 − I2
Current source in a mesh
fixes that mesh current

Method

  1. Draw a clockwise current in every mesh (window) of the circuit.
  2. Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
  3. In a resistor shared by two meshes the current is the difference of the two mesh currents.
  4. Solve the equations; a negative result means the current flows the other way.

Superposition

Superposition
x = x1 + x2 + … (one source at a time)
Switched-off voltage source
short circuit (0 V)
Switched-off current source
open circuit (0 A)
Voltage divider
V2 = V·R2/(R1 + R2)
Current divider
I1 = I·R2/(R1 + R2)

Method

  1. Count the independent sources. You will solve one simpler circuit per source.
  2. Keep one source and switch off the others: a voltage source becomes a short circuit (a wire), a current source becomes an open circuit (a gap).
  3. Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
  4. Repeat for every source, then add the contributions with their signs.

Thévenin & Norton

Thévenin voltage
Vth = Vopen circuit
Thévenin resistance
Rth: all sources off, seen from a–b
Norton current
IN = Vth/Rth
Load current
IL = Vth/(Rth + RL)
Maximum power transfer
RL = Rth, Pmax = Vth²/(4Rth)

Method

  1. Remove the load and mark the two terminals a and b.
  2. Find the open-circuit voltage between a and b: this is Vth.
  3. Switch off all independent sources and find the resistance seen from a–b: this is Rth.
  4. Put the load back on the equivalent circuit: Vth in series with Rth.
  5. For the Norton equivalent, convert: IN = Vth/Rth, with RN = Rth in parallel.

Op-amp circuits

Inverting amplifier
Vout = −(Rf/Rin)·Vin
Non-inverting amplifier
Vout = (1 + Rf/R1)·Vin
Summing amplifier
Vout = −Rf(V1/R1 + V2/R2)
Ideal op-amp
V+ = V−, Iin = 0

RC & RL transients

Time constant
τ = RC | τ = L/R
Response
x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ
Capacitor at DC
open circuit; vC cannot jump
Inductor at DC
short circuit; iL cannot jump
Settled
t ≈ 5τ

Method

  1. Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
  2. Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
  3. Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
  4. Use x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ.

2Practice problems

1.Series & parallel resistorsMedium
R1​ = 6 ΩR3​ = 2 Ω+−V1​24 VR2​4 ΩAR4​2 ΩB

Find the equivalent resistance Req seen by the source V1.

V1 = 24 VR1 = 6 ΩR2 = 4 ΩR3 = 2 ΩR4 = 2 Ω
Req = Ω
2.Series & parallel resistorsHard
R1​ = 20 ΩR3​ = 30 ΩR5​ = 4 Ω+−V1​36 VR2​2 ΩAR4​6 ΩBR6​4 Ω

Find the equivalent resistance Req seen by the source V1.

V1 = 36 VR1 = 20 ΩR2 = 2 ΩR3 = 30 ΩR4 = 6 ΩR5 = 4 ΩR6 = 4 Ω
Req = Ω
3.Nodal analysisMedium
R1​20 Ω+−V1​15 VR2​5 Ω−+V2​24 VR3​3 Ω+−V3​6 VI3​A

Three branches are connected between node A and ground, as shown. Use nodal analysis to find the current I3 in branch 3, flowing up into node A in the direction of the arrow.

V1 = 15 VR1 = 20 ΩV2 = 24 VR2 = 5 ΩV3 = 6 VR3 = 3 Ω
I3 = A
4.Nodal analysisHard
R1​ = 25 ΩR3​ = 3 Ω+−V1​9 VR2​4 ΩAR4​5 ΩB0.05·VA​

The diamond is a voltage-controlled current source that pushes a current of 0.05·VA (in A, with VA in V) up into node B. Use nodal analysis to find the node voltage VA (measured from ground).

V1 = 9 VR1 = 25 ΩR2 = 4 ΩR3 = 3 ΩR4 = 5 Ωg = 0.05 A/V
VA = V
5.Mesh analysisMedium
R1​ = 4 ΩR2​ = 10 ΩIR2​+−V1​9 VR3​5 ΩA+−V2​9 VI1​I2​

Use mesh analysis to find the current IR2 through R2 in the direction of the arrow.

V1 = 9 VV2 = 9 VR1 = 4 ΩR2 = 10 ΩR3 = 5 Ω
IR2 = A
6.Mesh analysisHard
R1​ = 4 ΩR2​ = 10 ΩR3​ = 4 Ω+−VR3​+−V1​24 VR4​25 ΩAR5​5 ΩB+−V2​12 VI1​I2​I3​

Use mesh analysis to find the voltage VR3 across R3 with the polarity shown.

V1 = 24 VV2 = 12 VR1 = 4 ΩR2 = 10 ΩR3 = 4 ΩR4 = 25 ΩR5 = 5 Ω
VR3 = V
7.SuperpositionMedium
R1​ = 10 ΩR3​ = 15 Ω+−V1​6 VR2​12 Ω+−VR2​AR4​10 ΩBIs​2.5 A

Use superposition to find the voltage VR2 across R2 with the polarity shown.

V1 = 6 VIs = 2.5 AR1 = 10 ΩR2 = 12 ΩR3 = 15 ΩR4 = 10 Ω
VR2 = V
8.SuperpositionHard
linear circuitV1​V2​Is​IRL​

A linear resistive circuit contains three independent sources: V1, V2 and Is. The current I through a load is measured three times. With all sources at their normal values, I = 3.5 mA. With V1 doubled and the other two unchanged, I = 7 mA. With the direction of Is reversed and the other two unchanged, I = 22.5 mA. Find the load current I when V2 is switched off and V1 and Is have their normal values.

normal: I = 3.5 mA2·V1: I = 7 mA−Is: I = 22.5 mA
I = mA
9.Thévenin & NortonMedium
R1​ = 20 ΩR3​ = 3 Ω+−V1​18 VR2​3 ΩAab

Find the Thévenin resistance Rth seen from terminals a–b.

V1 = 18 VR1 = 20 ΩR2 = 3 ΩR3 = 3 Ω
Rth = Ω
10.Thévenin & NortonHard
R1​ = 8 ΩR3​ = 20 Ω+−V1​18 VR2​5 ΩAIs​5 ABab

Find the Norton current IN: the current that flows from a to b if the terminals are shorted.

V1 = 18 VIs = 5 AR1 = 8 ΩR2 = 5 ΩR3 = 20 Ω
IN = A
11.Op-amp circuitsMedium
−+Vout​Rf​ = ?R1​ = 2 kΩVin​

Choose the feedback resistor Rf that gives this non-inverting amplifier a voltage gain of 11.

R1 = 2 kΩAv = 11
Rf = kΩ
12.Op-amp circuitsHard
−+Vout​C = 1 µFVin​ = −2.9 VR = 1 kΩ

The integrator uses an ideal op-amp with ±15 V supplies; its output saturates at ±12 V. The capacitor is uncharged when a constant input Vin = −2.9 V is applied at t = 0. Find the output voltage Vout at t = 0.5 ms.

Vin = −2.9 VR = 1 kΩC = 1 µFVsat = ±12 Vt = 0.5 ms
Vout = V
13.RC & RL transientsMedium
R = 20 Ωt = 0L150 mHi(t)

The inductor carries a steady current of 2 A. At t = 0 the switch closes the loop through R and the source is removed, so the inductor current decays through R. Find the current i at t = 10 ms.

i(0) = 2 AR = 20 ΩL = 150 mHt = 10 ms
i = A
14.RC & RL transientsHard
+−Vs​15 Vopens at t = 0R1​ = 4.7 kΩR2​ = 6.8 kΩC220 µF+−vC​

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage vC at t = 500 ms.

Vs = 15 VR1 = 4.7 kΩR2 = 6.8 kΩC = 220 µFt = 500 ms
vC = V

3Solutions

1.Series & parallel resistors

Answer: Req = 8 Ω

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 6 ΩR3​ = 2 Ω+−V1​24 VR2​4 ΩAR4​2 ΩB

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R3 + R4 = 2 + 2 = 4 Ω
R2 ∥ 4 = 4·4/(4+4) = 2 Ω
Req = R1 + 2 = 6 + 2 = 8 Ω

Answer

Req = 8 Ω
2.Series & parallel resistors

Answer: Req = 21.89 Ω

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 20 ΩR3​ = 30 ΩR5​ = 4 Ω+−V1​36 VR2​2 ΩAR4​6 ΩBR6​4 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R5 + R6 = 4 + 4 = 8 Ω
R4 ∥ 8 = 3.429 Ω
R3 + 3.429 = 33.43 Ω
R2 ∥ 33.43 = 1.887 Ω
Req = R1 + 1.887 = 20 + 1.887 = 21.89 Ω

Answer

Req = 21.89 Ω
3.Nodal analysis

Answer: I3 = 3.171 A

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

KCL at node A

Take the bottom wire as ground. Each branch carries (VA − Vk)/Rk away from A, where Vk is the voltage of its source (zero for a plain resistor). The currents leaving A add up to zero.

(VA − 15)/20 + (VA − (−24))/5 + (VA − 6)/3 = 0

Solve for VA

VA = (15/20 + (−24)/5 + 6/3) / (1/20 + 1/5 + 1/3) = −2.05/0.5833 = −3.514 V

Branch current

The arrow points up through the branch, into node A.

I3 = (6 − VA)/R3 = (6 − (−3.514))/3 = 3.171 A

Answer

I3 = 3.171 A

A negative result means the real current flows against the arrow.

4.Nodal analysis

Answer: VA = 0.9381 V

Nodal analysis: choose a reference node, label the unknown node voltages, write Kirchhoff’s current law at each unknown node using Ohm’s law (I = ΔV/R), and solve the resulting equations.

KCL at node A

Currents leaving A through the resistors equal the current pushed into A by sources.

(VA − V1)/R1 + VA/R2 + (VA − VB)/R3 = 0
0.6233·VA − 0.3333·VB = 0.36

KCL at node B

The dependent source pushes g·VA into node B. Treat it like an independent source, then move the VA term to the left-hand side.

(VB − VA)/R3 + VB/R4 = 0.05·VA
−0.3833·VA + 0.5333·VB = 0

Solve

Δ = (0.6233)(0.5333) − (−0.3333)(−0.3833) = 0.2047
VA = Δ1/Δ = 0.192/0.2047 = 0.9381 V
VB = Δ2/Δ = 0.138/0.2047 = 0.6743 V

Answer

VA = 0.9381 V
5.Mesh analysis

Answer: IR2 = −0.3273 A

Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.

R1​ = 4 ΩR2​ = 10 Ω+−V1​9 VR3​5 ΩA+−V2​9 VI1​I2​

Assign mesh currents

Draw a clockwise current in each window: I1, I2. A resistor shared by two meshes carries the difference of their currents.

KVL around mesh 1

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R1·I1 + R3(I1 − I2) = V1
4·I1 + 5(I1 − I2) = 9
9·I1 − 5·I2 = 9

KVL around mesh 2

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R2·I2 + R3(I2 − I1) = −V2
10·I2 + 5(I2 − I1) = (−9)
−5·I1 + 15·I2 = −9

Solve

Δ = (9)(15) − (−5)(−5) = 110
I1 = Δ1/Δ = 90/110 = 0.8182 A
I2 = Δ2/Δ = −36/110 = −0.3273 A

Answer

IR2 = I2 = −0.3273 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: VR3 = −3.345 V

Mesh analysis: give each window of a planar circuit its own loop current, write Kirchhoff’s voltage law around each loop (sum of drops = sum of rises), and solve. Branch currents are the mesh current, or the difference of two mesh currents on a shared branch.

R1​ = 4 ΩR2​ = 10 ΩR3​ = 4 Ω+−V1​24 VR4​25 ΩAR5​5 ΩB+−V2​12 VI1​I2​I3​

Assign mesh currents

Draw a clockwise current in each window: I1, I2, I3. A resistor shared by two meshes carries the difference of their currents.

KVL around mesh 1

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R1·I1 + R4(I1 − I2) = V1
4·I1 + 25(I1 − I2) = 24
29·I1 − 25·I2 = 24

KVL around mesh 2

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R2·I2 + R4(I2 − I1) + R5(I2 − I3) = 0
10·I2 + 25(I2 − I1) + 5(I2 − I3) = 0
−25·I1 + 40·I2 − 5·I3 = 0

KVL around mesh 3

Going clockwise, the voltage drops across resistors equal the source voltage rises (− to + in the direction of travel).

R3·I3 + R5(I3 − I2) = −V2
4·I3 + 5(I3 − I2) = (−12)
−5·I2 + 9·I3 = −12

Solve

Δ = det of the coefficient matrix = 4090
I1 = Δ1/Δ = 6540/4090 = 1.599 A
I2 = Δ2/Δ = 3660/4090 = 0.8949 A
I3 = Δ3/Δ = −3420/4090 = −0.8362 A

Answer

IR3 = I3 = −0.8362 A
VR3 = IR3·R3 = −0.8362·4 = −3.345 V

A negative mesh current flows anticlockwise.

7.Superposition

Answer: VR2 = 7.164 V

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

R1​ = 10 ΩR3​ = 15 Ω+−V1​6 VR2​12 ΩAR4​10 ΩBOPEN

Case 1 · V1 acting alone

Iₛ becomes an open circuit, so R₃ and R₄ form one series branch from A to ground, in parallel with R₂.

R3+R4 = 15+10 = 25 Ω
R2 ∥ (R3+R4) = 12·25/(12+25) = 8.108 Ω
VA = 6·8.108/(10+8.108) = 2.687 V
VB = VA·R4/(R3+R4) = 2.687·10/25 = 1.075 V
VR2′ = VA − 0 = 2.687 − 0 = 2.687 V
R1​ = 10 ΩR3​ = 15 ΩSHORTR2​12 ΩAR4​10 ΩBIs​2.5 A

Case 2 · Is acting alone

V₁ becomes a short, so R₁ is in parallel with R₂. Iₛ splits at B between R₄ and the path through R₃.

R1 ∥ R2 = 10·12/(10+12) = 5.455 Ω
Req = R4 ∥ (R3+R1∥R2) = 10 ∥ 20.45 = 6.716 Ω
VB = Is·Req = 2.5·6.716 = 16.79 V
VA = VB·(R1∥R2)/(R3+R1∥R2) = 16.79·5.455/20.45 = 4.478 V
VR2″ = VA − 0 = 4.478 − 0 = 4.478 V

Add the contributions

VR2 = VR2′ + VR2″
VR2 = 2.687 + 4.478 = 7.164 V

A negative result means the real polarity is opposite to the one marked.

8.Superposition

Answer: I = −6 mA

Superposition: in a linear circuit, any voltage or current is the sum of the responses to each independent source acting alone. To switch a source off, replace an ideal voltage source with a short circuit and an ideal current source with an open circuit. Resistors stay where they are.

Write the measurements with contributions

Each contribution is proportional to its own source.

I₁ + I₂ + I₃ = 3.5
2·I₁ + I₂ + I₃ = 7
I₁ + I₂ − I₃ = 22.5

Solve for the contributions

I₁ = 7 − 3.5 = 3.5 mA
I₃ = (3.5 − 22.5)/2 = −9.5 mA
I₂ = 3.5 − 3.5 − (−9.5) = 9.5 mA

Switch V₂ off

I = I₁ + I₃ = 3.5 + (−9.5) = −6 mA

Answer

I = −6 mA
9.Thévenin & Norton

Answer: Rth = 5.609 Ω

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 20 ΩR3​ = 3 ΩSHORTR2​3 ΩAabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 20·3/(20+3) = 2.609 Ω
Rth = (R1 ∥ R2) + R3 = 2.609 + 3 = 5.609 Ω
10.Thévenin & Norton

Answer: IN = −4.7 A

Thévenin’s theorem: any linear two-terminal network behaves like a single voltage source Vth in series with a resistance Rth. Norton’s theorem gives the same network as a current source IN = Vth/Rth in parallel with Rth.

R1​ = 8 ΩR3​ = 20 ΩSHORTR2​5 ΩAOPENBabRth​

Find Rth with every source switched off

Replace voltage sources with shorts and current sources with opens, then look into the circuit from a–b.

R1 ∥ R2 = 8·5/(8+5) = 3.077 Ω
Rth = (R1 ∥ R2) + R3 = 3.077 + 20 = 23.08 Ω
R1​ = 8 ΩR3​ = 20 Ω+−V1​18 VR2​5 ΩAIs​5 ABab+−Vth​

Find Vth, the open-circuit voltage

With a–b open, the only path for Iₛ is through R₃ into node A. Find V_A by superposition of V₁ and Iₛ, then add the drop across R₃.

VA = V1·R2/(R1+R2) + Is·(R1∥R2) = 6.923 + (−5)·3.077 = −8.462 V
Vth = VB = VA + Is·R3 = −8.462 + (−5)·20 = −108.5 V
IN​4.7 ARth​23.08 Ωab

Norton equivalent

A Norton source is the same circuit seen as a current source in parallel with R_th. Its current is the short-circuit current at a–b.

IN = Vth/Rth = −108.5/23.08 = −4.7 A
11.Op-amp circuits

Answer: Rf = 20 kΩ

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Gain of a non-inverting amplifier

V₋ = V₊ = Vin, and the current through R₁ also flows through Rf, so Vout = Vin·(1 + Rf/R₁).

Av = 1 + Rf/R1
Rf = (11 − 1)·2 = 20 kΩ

Answer

Rf = 20 kΩ
12.Op-amp circuits

Answer: Vout = 1.45 V

Ideal op-amp rules: with negative feedback, the op-amp drives its output so that its two inputs sit at the same voltage (V₋ = V₊), and no current flows into either input. Writing KCL at the inverting input with these two facts gives the output.

Input current and ramp rate

kΩ times µF gives milliseconds.

RC = 1 kΩ · 1 µF = 1 ms
ramp rate = −Vin/(RC) = −(−2.9)/1 = 2.9 V/ms

Output at the given time

The ideal value lies inside ±Vsat, so the integrator has not clipped yet.

Vout = −Vin·t/(RC) = −(−2.9)·0.5/1 = 1.45 V

Answer

Vout = 1.45 V
13.RC & RL transients

Answer: i = 0.5272 A

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Time constant

τ = L/R = 150 mH / 20 Ω = 7.5 ms

Decay of the current

i(t) = 2·e−10/7.5 = 0.5272 A

Answer

i = 0.5272 A
14.RC & RL transients

Answer: vC = 6.35 V

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

vC(0) = 15·6.8/(4.7 + 6.8) = 8.87 V

Time constant after the switch opens

τ = R2·C = 6.8·220 = 1496 ms

Discharge

vC(t) = 8.87·e−500/1496 = 6.35 V

Answer

vC = 6.35 V

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