Network analysis (circuit theory) is the set of methods used to find the voltage and current in every part of an electrical circuit: Kirchhoff's laws, nodal and mesh analysis, and theorems such as Thevenin, Norton, superposition and maximum power transfer. Below, one circuit is solved several ways with every number checked, followed by dependent sources, AC, transients and two-port networks. FLN Tutors (Freelancer Network) also offers one-to-one help. Matched to a verified specialist who knows your course, your university's marking criteria, and your deadline.
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Network analysis finds the unknown voltages, currents and powers in a circuit of sources, resistors, inductors and capacitors. It is usually the first core course in electrical and electronics engineering, also called *circuit theory* or *circuit analysis*. (The phrase is also used for computer and social networks; for routing and subnetting see computer networks help.)
| Tool | What it gives you | Best used when |
|---|---|---|
| KCL and KVL | Current balance at nodes, voltage balance round loops | Always; every method rests on them |
| Nodal analysis | All node voltages | Few nodes, current sources |
| Mesh analysis | All mesh currents | Few meshes, voltage sources |
| Thevenin / Norton | A two-component equivalent seen by a load | The load changes, or you need one branch |
| Superposition | The effect of each source alone | Several independent sources |
| Maximum power transfer | The load that draws most power | Matching a load to a source |
| Two-port parameters | Input-output model of a block | Amplifiers, filters, cascaded stages |
All of these assume a linear circuit. Diodes and transistors are handled in analog electronics once a linear model is chosen.
Kirchhoff's current law (KCL): the algebraic sum of currents at a node is zero, because charge is conserved. Kirchhoff's voltage law (KVL): the algebraic sum of voltages round any closed loop is zero, because energy is conserved.
Use the passive sign convention: label each element so current enters its + terminal; then p = v x i is power absorbed, and a negative value means the element is delivering power. For KVL, walk round the loop one way and add each drop (+ if you enter at the + sign). For KCL, count currents leaving as positive at every node in the question.
Finish with a power balance: power delivered by sources must equal power absorbed by everything else.
Nodal analysis writes one KCL equation per unknown node voltage: *n* essential nodes need *n* - 1 equations, as one node is the 0 V reference.
Worked example (used again below). A 10 V source V1 connects through R1 = 2 ohm to node A. From A, R2 = 4 ohm goes to ground and R3 = 4 ohm goes to the + terminal of a 6 V source V2, whose - terminal is grounded.
KCL at A: (V_A - 10)/2 + V_A/4 + (V_A - 6)/4 = 0. Multiply by 4: 4V_A - 26 = 0, so V_A = 6.5 V.
| Branch | Current | Direction |
|---|---|---|
| R1 (2 ohm) | (10 - 6.5)/2 = 1.75 A | From V1 into A |
| R2 (4 ohm) | 6.5/4 = 1.625 A | A to ground |
| R3 (4 ohm) | (6.5 - 6)/4 = 0.125 A | A into the + terminal of V2 |
Checks: 1.75 = 1.625 + 0.125. V1 delivers 10 x 1.75 = 17.5 W; the resistors absorb 6.125 + 10.5625 + 0.0625 = 16.75 W and V2 absorbs 6 x 0.125 = 0.75 W, as it is being charged. 16.75 + 0.75 = 17.5 W.
Mesh analysis writes one KVL equation per mesh (a loop with no loop inside it) and works only for planar circuits. Give each mesh a clockwise current; a resistor shared by two meshes carries their difference. A current source shared by two meshes needs a supermesh.
The same circuit by mesh. Mesh 1 holds V1, R1 and R2; mesh 2 holds R2, R3 and V2.
The determinant is 48 - 16 = 32. By Cramer's rule i1 = (80 - 24)/32 = 1.75 A and i2 = (-36 + 40)/32 = 0.125 A; R2 carries i1 - i2 = 1.625 A. Both methods agree, which is the best proof both are right.
| Situation | Choose |
|---|---|
| Fewer nodes (minus reference) than meshes | Nodal |
| Fewer meshes than nodes | Mesh |
| Many current sources, op-amps or a non-planar circuit | Nodal |
| Many voltage sources in series paths | Mesh |
If a question names the method, use it.
Thevenin's theorem: any linear two-terminal network can be replaced by a voltage source V_TH in series with a resistance R_TH. Norton's theorem: the same network as a current source I_N = V_TH / R_TH in parallel with R_TH.
Worked example. Take the left part of the circuit (10 V, 2 ohm, 4 ohm) seen from node A.
A 4 ohm load now draws 6.667/5.333 = 1.25 A with no re-solving. Reconnect the R3 and V2 branch and the current is (6.667 - 6)/5.333 = 0.125 A, matching nodal and mesh. Swapping between the two forms is a source transformation: 10 V behind 2 ohm equals 5 A in parallel with 2 ohm.
A dependent source takes its value from a voltage or current elsewhere in the circuit, as a transistor model does. Never switch it off when finding R_TH.
Worked example. A 10 V source drives current I_x through 2 ohm into node a. From a, 4 ohm goes to ground, and a dependent current source pushes 0.5 I_x into a.
Combining resistors alone gives 2 parallel 4 = 1.333 ohm, which is wrong. Examiners set dependent-source questions to catch exactly that.
Superposition: in a linear circuit, each voltage or current is the sum of the values caused by each independent source acting alone.
Power does not superpose. The partial powers in R2 add to 6.25 + 0.5625 = 6.8125 W, but the true value is 6.5 squared / 4 = 10.5625 W. Add voltages or currents, then compute power once.
Maximum power transfer: a load draws most power when its resistance equals R_TH, and P_max = V_TH squared / (4 R_TH).
| Load on the 6.667 V, 1.333 ohm source | Load power |
|---|---|
| 4 ohm | 1.25 squared x 4 = 6.25 W |
| 1.333 ohm (matched) | 44.44 / 5.333 = 8.333 W |
Efficiency when matched is only 50 per cent, so power systems avoid it while signal circuits use it. In AC, the load should be the complex conjugate of the source impedance.
AC steady state. Replace each element by its impedance (R, jwL, 1/jwC) and each source by a phasor; every method above then works with complex numbers.
Series resonance occurs at f0 = 1/(2 pi root(LC)). With L = 10 mH, C = 1 uF and R = 10 ohm: f0 = 1,591.55 Hz, Q = (1/R) root(L/C) = 10 and bandwidth f0/Q = 159.15 Hz.
First-order transients follow x(t) = x(final) + [x(initial) - x(final)] e^(-t/tau).
| Circuit | Time constant | Example |
|---|---|---|
| RC | tau = RC | 10 kohm, 100 uF: tau = 1 s; from 12 V, v(1 s) = 7.585 V |
| RL | tau = L/R | 0.5 H, 10 ohm: tau = 0.05 s; final 1.2 A, i(0.1 s) = 1.038 A |
Capacitor voltage and inductor current cannot jump, so carry them across the switching instant. Second-order and Laplace-domain analysis is covered in signals and systems help.
A two-port network has an input pair and an output pair of terminals, described by four parameters.
| Set | Defines | Typical use |
|---|---|---|
| Z | V1, V2 from I1, I2 | Series connections |
| Y | I1, I2 from V1, V2 | Parallel connections |
| h | V1, I2 from I1, V2 | Transistor models |
| ABCD | V1, I1 from V2, -I2 | Cascaded stages, lines |
Worked example. A T network has 2 ohm in the input arm, 4 ohm shunt and 6 ohm in the output arm.
z12 = z21 shows the network is reciprocal, and AD - BC = 3.75 - 2.75 = 1 confirms it.
From the circuit questions our tutors see most often:
| Level | Where it appears |
|---|---|
| School | Kirchhoff's rules and the Wheatstone bridge in CBSE Class 12 Physics and A-level DC circuits |
| First year | KVL, KCL, nodal, mesh, theorems, RC and RL transients |
| Second year | AC power, resonance, two-ports, Laplace-domain circuits |
| Exams | GATE EE and EC network theory, PSU papers |
Tools. LTspice, Multisim, PSpice and Proteus check a hand answer; MATLAB solves the matrix form.
Country notes. Indian syllabi (AKTU, VTU, Anna University, KTU) call the course *Network Analysis* or *Circuit Theory*, with Van Valkenburg as the standard text. UK and Australian modules say *Circuit Analysis*; US courses say *Circuits I* and *II*. Free notes from MIT OpenCourseWare 6.002 are a reliable second source.
A tutor can work through your circuit with you: choosing nodal or mesh, setting up equations, finding V_TH and R_TH and checking the answer a second way. For the wider subject, see electrical engineering help, which also covers star-delta and three-phase power. For related topics, see control systems help, electrical measurements help for bridges and instruments, and circuit design and analysis help for design projects.
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Network analysis is the process of finding the voltages, currents and powers in every element of an electrical circuit. It uses Kirchhoff's current and voltage laws, systematic methods such as nodal and mesh analysis, and theorems such as Thevenin, Norton, superposition and maximum power transfer to reduce complex circuits to simple ones.
Yes, in electrical engineering the two names mean the same course. Indian universities usually say network analysis or network theory, while UK, US and Australian universities more often say circuit analysis or electric circuits. The phrase network analysis is also used for computer networks and social networks, which are different subjects.
Kirchhoff's current law states that the algebraic sum of currents entering any node is zero, because charge is conserved. Kirchhoff's voltage law states that the algebraic sum of voltages round any closed loop is zero, because energy is conserved. Together they give enough equations to solve any linear circuit.
Count the equations each method needs. Mesh analysis needs one per mesh, nodal analysis one per node excluding the reference. Choose mesh when there are fewer meshes and many voltage sources; choose nodal when there are fewer nodes, many current sources or a non-planar circuit that mesh analysis cannot handle.
A supernode is formed when a voltage source sits between two non-reference nodes. You enclose both nodes and the source in one boundary and write a single KCL equation for everything leaving it, then add a second equation stating that the difference between the two node voltages equals the source voltage.
Remove the load. The Thevenin voltage is the open-circuit voltage across the load terminals. The Thevenin resistance is the resistance seen from those terminals with independent voltage sources shorted and current sources opened. If the circuit has dependent sources, use the open-circuit voltage divided by the short-circuit current instead.
Keep the dependent source active. Either divide the open-circuit voltage by the short-circuit current, or turn off independent sources only, apply a test source of 1 ampere or 1 volt at the terminals, and divide test voltage by test current. Simply combining resistors gives a wrong answer whenever a dependent source is present.
They describe the same circuit. The Norton resistance equals the Thevenin resistance, and the Norton current equals the Thevenin voltage divided by that resistance, which is also the short-circuit current at the terminals. Converting between them is a source transformation, so you can use whichever form makes the next step easier.
Power depends on the square of current or voltage, so it is not a linear quantity. Adding the powers caused by each source separately leaves out the cross term. Superpose the currents or voltages first, add them to get the total, and only then calculate power from the total value.
The theorem states that a load receives the most power from a linear source when the load resistance equals the source's Thevenin resistance. The maximum power is the Thevenin voltage squared divided by four times the Thevenin resistance. Efficiency at that point is only 50 per cent, as the source resistance dissipates equal power.
In an AC circuit the load impedance must equal the complex conjugate of the source impedance: the same resistance with reactance of opposite sign. The reactances then cancel and the maximum power equals the rms Thevenin voltage squared divided by four times the source resistance. A purely resistive load gives less.
The time constant is RC for a resistor-capacitor circuit and L divided by R for a resistor-inductor circuit. After one time constant a charging response reaches about 63.2 per cent of its final change, and after five time constants it reaches about 99.3 per cent, which is usually treated as steady state.
Two-port parameters describe a network by how its input and output voltages and currents relate. The common sets are impedance (Z), admittance (Y), hybrid (h) and transmission (ABCD) parameters. Each set suits a different connection: Z for series, Y for parallel, h for transistor models and ABCD for cascaded stages.
Van Valkenburg's Network Analysis is the standard text in Indian universities. Hayt, Kemmerly and Durbin's Engineering Circuit Analysis and Alexander and Sadiku's Fundamentals of Electric Circuits are widely used in UK and US courses. Use the one your syllabus lists, as notation differs.
Yes. A one-to-one tutor can work through nodal, mesh, theorem, transient and two-port problems with you, show where your equations go wrong and check answers by a second method. You complete and submit your own assessed work in line with your college or university's academic integrity rules.
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