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

Method

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

Nodal analysis

Kirchhoff’s current law: ΣI_in = ΣI_out
Branch current: I = (V_a − V_b)/R
Reference node: V_ground = 0

Method

Choose a reference node (ground) and label the other node voltages.
Write Kirchhoff’s current law at every unknown node: currents leaving through each branch equal the currents entering.
Express every branch current as (V_a − V_b)/R.
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

Draw a clockwise current in every mesh (window) of the circuit.
Apply Kirchhoff’s voltage law around each mesh: the sum of voltage drops is zero.
In a resistor shared by two meshes the current is the difference of the two mesh currents.
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

Count the independent sources. You will solve one simpler circuit per source.
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).
Solve the simpler circuit for the voltage or current you need, keeping track of its direction.
Repeat for every source, then add the contributions with their signs.

Thévenin & Norton

Thévenin voltage: V_th = V_{open circuit}
Thévenin resistance: R_th: all sources off, seen from a–b
Norton current: I_N = V_th/R_th
Load current: I_L = V_th/(R_th + R_L)
Maximum power transfer: R_L = R_th, P_max = V_th²/(4R_th)

Method

Remove the load and mark the two terminals a and b.
Find the open-circuit voltage between a and b: this is V_th.
Switch off all independent sources and find the resistance seen from a–b: this is R_th.
Put the load back on the equivalent circuit: V_th in series with R_th.
For the Norton equivalent, convert: I_N = V_th/R_th, with R_N = R_th in parallel.

Op-amp circuits

Inverting amplifier: V_out = −(R_f/R_in)·V_in
Non-inverting amplifier: V_out = (1 + R_f/R₁)·V_in
Summing amplifier: V_out = −R_f(V₁/R₁ + V₂/R₂)
Ideal op-amp: V₊ = V₋, I_in = 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; v_C cannot jump
Inductor at DC: short circuit; i_L cannot jump
Settled: t ≈ 5τ

Method

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

2Practice problems

1.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 15 VR₁ = 3 ΩR₂ = 4 ΩR₃ = 4 ΩR₄ = 12 Ω

R_eq = Ω

2.Series & parallel resistorsHard

Find the current I_R2 through R₂.

V₁ = 6 VR₁ = 15 ΩR₂ = 4 ΩR₃ = 3 ΩR₄ = 15 ΩR₅ = 10 ΩR₆ = 20 Ω

I_R2 = A

3.Nodal analysisMedium

Use nodal analysis to find the node voltage V_A (measured from ground).

V₁ = 6 VI_s = 5 AR₁ = 2 ΩR₂ = 4 ΩR₃ = 15 ΩR₄ = 5 Ω

V_A = V

4.Nodal analysisHard

Use nodal analysis to find the current I_R2 through R₂ in the direction of the arrow.

V₁ = 9 VI_s = 3 AV₂ = 12 VR₁ = 6 ΩR₂ = 10 ΩR₃ = 5 ΩR₄ = 12 ΩR₅ = 6 Ω

I_R2 = A

5.Mesh analysisMedium

Use mesh analysis to find the current I_R2 through R₂ in the direction of the arrow.

V₁ = 30 VV₂ = 15 VR₁ = 10 ΩR₂ = 15 ΩR₃ = 5 Ω

I_R2 = A

6.Mesh analysisHard

Use mesh analysis to find the mesh current I₂ (clockwise).

V₁ = 12 VV₂ = 5 VR₁ = 3 ΩR₂ = 30 ΩR₃ = 3 ΩR₄ = 12 ΩR₅ = 10 Ω

I₂ = A

7.SuperpositionMedium

Use superposition to find the current I_R1 through R₁ in the direction of the arrow.

V₁ = 30 VI_s = 3 AR₁ = 12 ΩR₂ = 4 ΩR₃ = 12 ΩR₄ = 4 Ω

I_R1 = A

8.SuperpositionHard

A linear resistive circuit contains three independent sources: V₁, V₂ and I_s. The current I through a load is measured three times. With all sources at their normal values, I = 15 mA. With V₁ doubled and the other two unchanged, I = 20 mA. With the direction of I_s reversed and the other two unchanged, I = 5 mA. Find the load current I when V₂ is switched off and V₁ and I_s have their normal values.

normal: I = 15 mA2·V₁: I = 20 mA−I_s: I = 5 mA

I = mA

9.Thévenin & NortonMedium

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

V₁ = 12 VR₁ = 15 ΩR₂ = 15 ΩR₃ = 4 Ω

R_th = Ω

10.Thévenin & NortonHard

Find the Thévenin voltage V_th seen from terminals a–b.

V₁ = 15 VI_s = 1 AR₁ = 5 ΩR₂ = 30 ΩR₃ = 2 Ω

V_th = V

11.Op-amp circuitsMedium

The op-amp is ideal and runs from ±15 V supplies. Find the output voltage V_out.

V_in = 1.1 VR₁ = 1 kΩR_f = 2 kΩ

V_out = V

12.Op-amp circuitsHard

Two ideal op-amp stages are connected in cascade: an inverting amplifier (R₁, R_f1) followed by a non-inverting amplifier (R₂ to ground, R_f2 in the feedback). Both run from ±15 V supplies. The input is V_in = 0.25 V. Find the output voltage V_out.

R₁ = 1 kΩR_f1 = 4.7 kΩR₂ = 4.7 kΩR_f2 = 4.7 kΩV_in = 0.25 V

V_out = V

13.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 10 ms, in mJ.

V_s = 6 VR = 10 ΩL = 47 mHt = 10 ms

W = mJ

14.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 10 V with the polarity shown. Find the capacitor voltage v_C at t = 60 ms.

V_s = 12 VR = 1 kΩC = 22 µFv_C(0) = 10 Vt = 60 ms

v_C = V

3Solutions

1.Series & parallel resistors

Answer: R_eq = 6.2 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.

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 = 4 + 12 = 16 Ω

R₂ ∥ 16 = 4·16/(4+16) = 3.2 Ω

R_eq = R₁ + 3.2 = 3 + 3.2 = 6.2 Ω

Answer

R_eq = 6.2 Ω

2.Series & parallel resistors

Answer: I_R2 = 0.2541 A

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 = 10 + 20 = 30 Ω

R₄ ∥ 30 = 10 Ω

R₃ + 10 = 13 Ω

R₂ ∥ 13 = 3.059 Ω

R_eq = R₁ + 3.059 = 15 + 3.059 = 18.06 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 6/18.06 = 0.3322 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

V_A = V₁ − I_R1·R₁ = 6 − 0.3322·15 = 1.016 V

I_R2 = V_A/R₂ = 1.016/4 = 0.2541 A

Answer

I_R2 = 0.2541 A

3.Nodal analysis

Answer: V_A = 5.313 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.

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V₁ = 6 V). Unknowns: V_A, V_B.

KCL at node A

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

V_A/R₂ + (V_A − V₁)/R₁ + (V_A − V_B)/R₃ = 0

V_A/4 + (V_A − 6)/2 + (V_A − V_B)/15 = 0

0.8167·V_A − 0.06667·V_B = 3

KCL at node B

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

V_B/R₄ + (V_B − V_A)/R₃ = I_s

V_B/5 + (V_B − V_A)/15 = 5

−0.06667·V_A + 0.2667·V_B = 5

Solve

Δ = (0.8167)(0.2667) − (−0.06667)(−0.06667) = 0.2133

V_A = Δ₁/Δ = 1.133/0.2133 = 5.313 V

V_B = Δ₂/Δ = 4.283/0.2133 = 20.08 V

Answer

V_A = 5.313 V

4.Nodal analysis

Answer: I_R2 = 0.7228 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.

Set up the nodes

Ground is the bottom wire. Known node voltages come straight from the sources (V₁ = 9 V, V₂ = 12 V). Unknowns: V_A, V_B.

KCL at node A

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

V_A/R₃ + (V_A − V₁)/R₁ + (V_A − V_B)/R₂ = 0

V_A/5 + (V_A − 9)/6 + (V_A − V_B)/10 = 0

0.4667·V_A − 0.1·V_B = 1.5

KCL at node B

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

V_B/R₅ + (V_B − V_A)/R₂ + (V_B − V₂)/R₄ = I_s

V_B/6 + (V_B − V_A)/10 + (V_B − 12)/12 = (−3)

−0.1·V_A + 0.35·V_B = −2

Solve

Δ = (0.4667)(0.35) − (−0.1)(−0.1) = 0.1533

V_A = Δ₁/Δ = 0.325/0.1533 = 2.12 V

V_B = Δ₂/Δ = −0.7833/0.1533 = −5.109 V

Answer

I_R2 = (V_A − V_B)/R₂ = (2.12 − −5.109)/10 = 0.7228 A

5.Mesh analysis

Answer: I_R2 = −0.2727 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.

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

10·I₁ + 5(I₁ − I₂) = 30

15·I₁ − 5·I₂ = 30

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

15·I₂ + 5(I₂ − I₁) = (−15)

−5·I₁ + 20·I₂ = −15

Solve

Δ = (15)(20) − (−5)(−5) = 275

I₁ = Δ₁/Δ = 525/275 = 1.909 A

I₂ = Δ₂/Δ = −75/275 = −0.2727 A

Answer

I_R2 = I₂ = −0.2727 A

A negative mesh current flows anticlockwise.

6.Mesh analysis

Answer: I₂ = 0.1658 A

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

3·I₁ + 12(I₁ − I₂) = 12

15·I₁ − 12·I₂ = 12

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

30·I₂ + 12(I₂ − I₁) + 10(I₂ − I₃) = 0

−12·I₁ + 52·I₂ − 10·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

3·I₃ + 10(I₃ − I₂) = (−5)

−10·I₂ + 13·I₃ = −5

Solve

Δ = det of the coefficient matrix = 6768

I₁ = Δ₁/Δ = 6312/6768 = 0.9326 A

I₂ = Δ₂/Δ = 1122/6768 = 0.1658 A

I₃ = Δ₃/Δ = −1740/6768 = −0.2571 A

Answer

I₂ = 0.1658 A

A negative mesh current flows anticlockwise.

7.Superposition

Answer: I_R1 = 2.132 A

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.

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 = 12+4 = 16 Ω

R₂ ∥ (R₃+R₄) = 4·16/(4+16) = 3.2 Ω

V_A = 30·3.2/(12+3.2) = 6.316 V

V_B = V_A·R₄/(R₃+R₄) = 6.316·4/16 = 1.579 V

I_R1′ = (V₁ − V_A)/R₁ = (30 − 6.316)/12 = 1.974 A

Case 2 · I_s 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 = 12·4/(12+4) = 3 Ω

R_eq = R₄ ∥ (R₃+R₁∥R₂) = 4 ∥ 15 = 3.158 Ω

V_B = I_s·R_eq = (−3)·3.158 = −9.474 V

V_A = V_B·(R₁∥R₂)/(R₃+R₁∥R₂) = −9.474·3/15 = −1.895 V

I_R1″ = (V₁ − V_A)/R₁ = (0 − −1.895)/12 = 0.1579 A

Add the contributions

I_R1 = I_R1′ + I_R1″

I_R1 = 1.974 + 0.1579 = 2.132 A

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

8.Superposition

Answer: I = 10 mA

Write the measurements with contributions

Each contribution is proportional to its own source.

I₁ + I₂ + I₃ = 15

2·I₁ + I₂ + I₃ = 20

I₁ + I₂ − I₃ = 5

Solve for the contributions

I₁ = 20 − 15 = 5 mA

I₃ = (15 − 5)/2 = 5 mA

I₂ = 15 − 5 − 5 = 5 mA

Switch V₂ off

I = I₁ + I₃ = 5 + 5 = 10 mA

Answer

I = 10 mA

9.Thévenin & Norton

Answer: R_th = 11.5 Ω

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

Find R_th 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 = 15·15/(15+15) = 7.5 Ω

R_th = (R₁ ∥ R₂) + R₃ = 7.5 + 4 = 11.5 Ω

10.Thévenin & Norton

Answer: V_th = 6.571 V

Find R_th 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 = 5·30/(5+30) = 4.286 Ω

R_th = (R₁ ∥ R₂) + R₃ = 4.286 + 2 = 6.286 Ω

Find V_th, 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.

V_A = V₁·R₂/(R₁+R₂) + I_s·(R₁∥R₂) = 12.86 + (−1)·4.286 = 8.571 V

V_th = V_B = V_A + I_s·R₃ = 8.571 + (−1)·2 = 6.571 V

11.Op-amp circuits

Answer: V_out = 3.3 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.

Inverting input follows the input

V₋ = V₊ = 1.1 V

I = V₋/R₁ = 1.1/1 = 1.1 mA

Output voltage

The current through R₁ also flows through R_f.

V_out = V₋ + I·R_f = 1.1 + 1.1·2 = 3.3 V

Answer

V_out = 3.3 V

12.Op-amp circuits

Answer: V_out = −2.35 V

Gain of each stage

A₁ = −R_f1/R₁ = −4.7/1 = −4.7

A₂ = 1 + R_f2/R₂ = 1 + 4.7/4.7 = 2

Stage by stage

V₁ = −4.7·0.25 = −1.175 V

V_out = 2·(−1.175) = −2.35 V

Answer

V_out = −2.35 V

13.RC & RL transients

Answer: W = 6.565 mJ

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 and final current

Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.

τ = L/R = 47 mH / 10 Ω = 4.7 ms

i(∞) = V_s/R = 6/10 = 0.6 A

Current at the given time

i(t) = 0.6·(1 − e^−10/4.7) = 0.5285 A

Stored energy

W = ½·L·i² = ½·47·0.5285² = 6.565 mJ

Answer

W = 6.565 mJ

14.RC & RL transients

Answer: v_C = 11.87 V

Initial value, final value, time constant

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

v_C(0) = 10 V, v_C(∞) = V_s = 12 V

τ = R·C = 1 kΩ · 22 µF = 22 ms

v_C(t) = 12 + (10 − 12)·e^−t/τ

Voltage at the given time

v_C(60) = 12 + (−2)·e^−60/22 = 11.87 V

Answer

v_C = 11.87 V

ExamProblems · examproblems.com/courses/circuit-analysis-1

Circuit Analysis 1: cram pack

1Formulas and methods

Series & parallel resistors

Method

Nodal analysis

Method

Mesh analysis

Method

Superposition

Method

Thévenin & Norton

Method

Op-amp circuits

RC & RL transients

Method

2Practice problems

3Solutions

Combine resistors from the far end

Answer

Combine resistors from the far end

Current from the source

Current through R₂

Answer

Set up the nodes

KCL at node A

KCL at node B

Solve

Answer

Set up the nodes

KCL at node A

KCL at node B

Solve

Answer

Assign mesh currents

KVL around mesh 1

KVL around mesh 2

Solve

Answer

Assign mesh currents

KVL around mesh 1

KVL around mesh 2

KVL around mesh 3

Solve

Answer

Case 1 · V1 acting alone

Case 2 · Is acting alone

Add the contributions

Write the measurements with contributions

Solve for the contributions

Switch V₂ off

Answer

Find Rth with every source switched off

Find Rth with every source switched off

Find Vth, the open-circuit voltage

Inverting input follows the input

Output voltage

Answer

Gain of each stage

Stage by stage

Answer

Time constant and final current

Current at the given time

Stored energy

Answer

Initial value, final value, time constant

Voltage at the given time

Answer

Case 1 · V₁ acting alone

Case 2 · I_s acting alone

Find R_th with every source switched off

Find R_th with every source switched off

Find V_th, the open-circuit voltage