Circuit analysis means finding the voltages, currents and powers in a circuit from its components. Ohm's law and Kirchhoff's two laws are enough to solve any linear DC circuit; the other methods are organised ways of applying them.
This guide explains each tool, then solves one two-source circuit by nodal analysis, mesh analysis and Thevenin's theorem, and checks the answer with a power balance. It finishes with RC circuits and the mistakes that cost marks.
The Tools That Solve Almost Any DC Circuit
- Ohm's law: V = IR for each resistor.
- Kirchhoff's current law (KCL): the currents into any node sum to zero.
- Kirchhoff's voltage law (KVL): the voltages around any closed loop sum to zero.
- Nodal or mesh analysis: systematic ways to write KCL or KVL equations and solve them.
- Thevenin's theorem: replaces a network with one source and one resistor, as seen by a load.
Ohm's Law, Series and Parallel
Ohm's law links voltage (V, in volts), current (I, in amperes) and resistance (R, in ohms): V = IR. Power in a resistor is P = VI = I²R = V²/R, in watts.
| Arrangement | Equivalent resistance | What is shared |
|---|---|---|
| Series | R = R₁ + R₂ + ... | The same current flows through each |
| Parallel | 1/R = 1/R₁ + 1/R₂ + ... | The same voltage appears across each |
| Two in parallel | R = R₁R₂ / (R₁ + R₂) | The same voltage appears across each |
Worked example: a voltage divider. A 9.0 V supply drives R₁ = 1.0 kΩ in series with R₂ = 2.0 kΩ. Find the current and the voltage across R₂.
- Total resistance: 1.0 + 2.0 = 3.0 kΩ.
- Current: I = 9.0 V / 3.0 kΩ = 3.0 mA.
- Voltage across R₂: V₂ = IR₂ = 3.0 mA × 2.0 kΩ = 6.0 V.
The divider formula gives the same result directly: V₂ = Vs × R₂ / (R₁ + R₂) = 9.0 × 2.0 / 3.0 = 6.0 V. Note that mA × kΩ gives volts, a handy unit pair.
Kirchhoff's Current and Voltage Laws
KCL comes from conservation of charge: charge cannot pile up at a node, so the current flowing in equals the current flowing out. KVL comes from conservation of energy: a charge that goes around a loop returns to the same potential.
Sign conventions are where most mistakes happen. Choose a direction for each unknown current before writing equations. If an answer comes out negative, the current simply flows the other way; do not go back and change the diagram.
The Example Circuit
The next three sections solve the same circuit, so you can compare methods. A 12 V source, with its positive terminal at the top, connects through R₁ = 2 Ω to a top node, V. A resistor R₂ = 4 Ω runs from that node to ground. A second branch runs from the node through R₃ = 4 Ω to the positive terminal of a 6 V source, whose negative terminal is grounded.
The goal is to find the node voltage V and the current in every resistor.
Method 1: Nodal Analysis
Nodal analysis applies KCL at each node whose voltage is unknown. Pick a ground node at 0 V, label the other node voltages, and write the current leaving the node through each branch as (Vnode − Vother end) / R.
KCL at node V, with all currents taken as leaving the node:
- (V − 12)/2 + V/4 + (V − 6)/4 = 0
- Multiply by 4: 2V − 24 + V + V − 6 = 0, so 4V = 30.
- V = 7.5 V.
Branch currents:
- Through R₁, from the 12 V source into the node: (12 − 7.5)/2 = 2.25 A.
- Through R₂, down to ground: 7.5/4 = 1.875 A.
- Through R₃, from the node into the 6 V source: (7.5 − 6)/4 = 0.375 A.
KCL check: 2.25 A in = 1.875 A + 0.375 A out.
Nodal analysis is usually the quickest method when a circuit has fewer nodes than loops, which is common in practice.
Method 2: Mesh Analysis
Mesh analysis applies KVL around each window of a planar circuit, using an imaginary mesh current in each loop. Where two meshes share a component, the current through it is the difference of the two mesh currents.
Let I₁ circulate clockwise in the left mesh (12 V source, R₁, R₂) and I₂ clockwise in the right mesh (R₂, R₃, 6 V source). R₂ carries I₁ − I₂ downwards.
- Left mesh: 12 = 2I₁ + 4(I₁ − I₂), so 6I₁ − 4I₂ = 12.
- Right mesh: 4I₂ + 6 + 4(I₂ − I₁) = 0, so −4I₁ + 8I₂ = −6.
- Double the first equation and add: 12I₁ − 8I₂ − 4I₁ + 8I₂ = 24 − 6, so 8I₁ = 18 and I₁ = 2.25 A.
- Substitute: 8I₂ = −6 + 4(2.25) = 3, so I₂ = 0.375 A.
- Current in R₂: I₁ − I₂ = 1.875 A, so V = 4 × 1.875 = 7.5 V.
Both methods agree, which is the best check there is.
Method 3: Thevenin's Theorem
Thevenin's theorem says any linear network, seen from two terminals, behaves like a single voltage source Vth in series with a resistance Rth. It is ideal when a question asks about one load that changes.
Treat R₂ as the load and find the Thevenin equivalent of the rest.
- Vth: remove R₂ and find the open-circuit node voltage. The same current flows from the 12 V source through R₁ and R₃ into the 6 V source: (12 − V)/2 = (V − 6)/4, so 24 − 2V = V − 6 and Vth = 10 V.
- Rth: replace both voltage sources with short circuits. R₁ and R₃ are then in parallel: 2 × 4 / (2 + 4) = 1.33 Ω.
- Reconnect the load: V = 10 × 4 / (4 + 1.33) = 7.5 V.
The same 7.5 V again. Maximum power would reach a load equal to Rth, giving Pmax = Vth² / (4Rth) = 100 / 5.33 = 18.75 W.
When a circuit contains dependent sources, you cannot find Rth by simply turning sources off. Apply a test source instead, or divide the open-circuit voltage by the short-circuit current.
Checking with a Power Balance
In any circuit, the power delivered by sources equals the power absorbed by everything else. A power balance is a strong final check on your currents.
| Element | Calculation | Power |
|---|---|---|
| 12 V source (delivers) | 12 V × 2.25 A | 27.0 W |
| R₁ | 2.25² × 2 | 10.125 W |
| R₂ | 1.875² × 4 | 14.0625 W |
| R₃ | 0.375² × 4 | 0.5625 W |
| 6 V source (absorbs, being charged) | 6 V × 0.375 A | 2.25 W |
Absorbed: 10.125 + 14.0625 + 0.5625 + 2.25 = 27.0 W, matching the 27.0 W delivered. Current enters the positive terminal of the 6 V source, so that source absorbs power, as a battery on charge does.
Beyond Resistors: RC Circuits and AC
Capacitors add time to the picture. When a capacitor charges through a resistor from a supply Vs, its voltage is VC = Vs(1 − e^(−t/τ)), where the time constant τ = RC. After one τ it reaches about 63% of Vs; after 5τ, over 99%.
In AC circuits, components have impedance: Z = R for a resistor, Z = jωL for an inductor and Z = 1/(jωC) for a capacitor, with ω = 2πf. Ohm's law and Kirchhoff's laws still apply, using complex numbers.
Worked example: RC low-pass filter. R = 1.0 kΩ and C = 100 nF. The cut-off frequency is fc = 1 / (2πRC) = 1 / (2π × 1.0 × 10³ × 100 × 10⁻⁹) = 1 / (6.28 × 10⁻⁴) = 1.6 kHz. At fc, the output amplitude is 1/√2, about 71%, of the input.
Common Circuit Analysis Mistakes
- Changing current directions part-way through instead of accepting a negative answer.
- Mixing sign conventions within one KVL equation.
- Treating components as parallel when they do not share both nodes.
- Forgetting unit prefixes: kΩ, mA, µF and nF.
- Turning off a current source by shorting it; it should be opened.
- Skipping a final check by a second method or a power balance.
How STEM Donkey Helps with Circuit Analysis
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Frequently Asked Questions
KCL says the currents into a node sum to zero, from conservation of charge. KVL says the voltages around a closed loop sum to zero, from conservation of energy.
Use whichever gives fewer equations. Nodal analysis needs one equation per unknown node voltage; mesh analysis needs one per window. Nodal analysis also works for non-planar circuits.
That the current flows opposite to the direction you assumed. Keep the answer and state the true direction.
With only independent sources, short voltage sources and open current sources, then find the resistance between the terminals. With dependent sources, use a test source or Voc divided by Isc.
Solve by a second method, check KCL at every node, or confirm that power delivered equals power absorbed.
τ = RC, in seconds when R is in ohms and C in farads. The capacitor reaches about 63% of its final voltage after one τ.
Yes. Send the brief and any files. Simulated results are compared with hand calculations and the differences explained.