Circuit Analysis¶

Exam-ready answer

Ohm's law: for a conductor whose temperature and other physical conditions remain constant, current is directly proportional to the applied potential difference. Thus

where \(V\) is voltage in volts (V), \(I\) is current in amperes (A), and \(R\) is resistance in ohms (\(\Omega\)). It does not describe a diode, transistor, arc lamp, or a conductor whose temperature changes appreciably because their \(V\)-\(I\) relation is non-linear.

Kirchhoff's current law (KCL): conservation of charge requires the algebraic sum of currents at a node to be zero. Taking currents entering as positive,

Kirchhoff's voltage law (KVL): conservation of energy requires the algebraic sum of all voltage rises and drops around a closed loop to be zero. For a chosen traversal direction, use a rise as positive and a drop as negative:

Application check: a \(12\,\text{V}\) source supplies series resistors \(R_1=2\,\Omega\) and \(R_2=4\,\Omega\). Ohm's law gives

At the junction, \(2\,\text{A}\) enters and \(2\,\text{A}\) leaves, so KCL gives \(2-2=0\). Traversing from the negative source terminal in the current direction, KVL gives

Hence Ohm's law determines each element drop, KCL checks current continuity, and KVL checks the complete energy balance. Together they are the basic laws used to analyse DC and lumped AC networks.

Practice target: 8–9 minutes; state all three laws, keep one sign convention, and verify one numerical loop.

Exam-ready answer

Mesh analysis is a systematic application of KVL in which one unknown current is assigned to every independent mesh of a planar circuit. Nodal analysis is a systematic application of KCL in which unknown node potentials are measured relative to a selected reference node. Both methods reduce a linear network to simultaneous algebraic equations.

Mesh method and sign rule¶

1. Confirm that the circuit is planar and identify its meshes.
2. Assign clockwise mesh currents \(I_1,I_2,\ldots\) in amperes (A); the choice is arbitrary but must remain consistent.
3. Write KVL for each mesh. A resistor belonging only to mesh 1 has drop \(R I_1\); a resistor shared by meshes 1 and 2 has drop \(R(I_1-I_2)\) in mesh 1.
4. Solve the simultaneous equations. A negative result means the true current is opposite to the assumed arrow.
5. If an ideal current source lies between two meshes, form a supermesh, write KVL around its perimeter, and add the source constraint such as \(I_2-I_1=I_s\).

Worked mesh example: two clockwise meshes share \(R_3=2\,\Omega\). The left mesh has \(R_1=2\,\Omega\) and a \(10\,\text{V}\) rise; the right has \(R_2=4\,\Omega\) and no source. KVL gives

Therefore \(I_1=3\,\text{A}\) and \(I_2=1\,\text{A}\); the shared-branch current is \(I_1-I_2=2\,\text{A}\) in the left-mesh direction. Substitution gives \(4(3)-2(1)=10\) and \(-2(3)+6(1)=0\), confirming both KVL equations.

Nodal method and sign rule¶

1. Select a reference node (\(0\,\text{V}\)), normally the node with most connections.
2. Label the remaining node voltages \(V_1,V_2,\ldots\) in volts (V).
3. Write each branch current as \((V_{node}-V_{other})/R\) and apply KCL, commonly as the sum of currents leaving the node equal to zero.
4. Solve the equations and then obtain any required branch current by Ohm's law.
5. If an ideal voltage source joins two non-reference nodes, enclose them in a supernode, write KCL around the boundary, and add the voltage constraint \(V_1-V_2=V_s\) with the shown polarity.

Worked nodal example: node \(V\) is connected to a fixed \(10\,\text{V}\) node through \(2\,\Omega\), and to ground through \(2\,\Omega\) and \(4\,\Omega\). Taking currents leaving \(V\),

Multiplying by \(4\,\Omega\) gives \(2(V-10)+2V+V=0\), hence \(V=4\,\text{V}\). The incoming current is \((10-4)/2=3\,\text{A}\) and outgoing currents are \(4/2=2\,\text{A}\) and \(4/4=1\,\text{A}\); KCL checks as \(3=2+1\,\text{A}\).

Thus the preferred method is the one producing fewer equations: mesh analysis is natural for voltage-driven planar networks, whereas nodal analysis is usually fastest for current-driven networks and electronic circuits.

Practice target: 16–18 minutes; draw the labeled circuit, write one equation per unknown, solve both examples, and finish with the comparison.

Exam-ready answer

Statement: in any linear bilateral network containing two or more independent sources, the voltage across or current through an element equals the algebraic sum of the responses produced by each independent source acting alone. Linearity means that scaling a source scales the response and individual responses can be added.

To retain only one independent source, replace every other ideal source by its zero-valued internal impedance. Dependent sources are controlled circuit elements and must remain active.

Procedure: (1) mark the required branch-current direction or voltage polarity; (2) activate one independent source and kill all others correctly; (3) calculate that signed partial response; (4) repeat for each source; and (5) add all partial responses algebraically.

Worked application: a node \(V\) is connected to a \(12\,\text{V}\) source through \(6\,\Omega\), to a \(6\,\text{V}\) source through \(3\,\Omega\), and to ground through a \(6\,\Omega\) load.

With the \(12\,\text{V}\) source alone, short the \(6\,\text{V}\) source. KCL gives

With the \(6\,\text{V}\) source alone, short the \(12\,\text{V}\) source:

Therefore \(V=V_1+V_2=6\,\text{V}\) and the load current is \(I_L=V/6=1\,\text{A}\). A direct all-source KCL check gives the same \(6\,\text{V}\).

Superposition applies to voltage and current, not directly to power because \(P=I^2R=V^2/R\) is non-linear; first sum the signed current or voltage, then calculate total power. The theorem is useful for multi-source bias, signal and fault calculations.

Practice target: 8–9 minutes; explicitly show both source-killing cases and add signed responses before computing power.

Exam-ready answer

Thevenin's theorem: any linear bilateral two-terminal network of sources and impedances can be replaced, as viewed from its load terminals, by an ideal voltage source \(V_{Th}\) in series with an equivalent resistance \(R_{Th}\). Here

the open-circuit terminal voltage. Norton's theorem: the same network can be replaced by an ideal current source \(I_N\) in parallel with \(R_N\), where

The two forms are source transformations:

Determination procedure¶

1. Remove the load \(R_L\) and mark terminal polarity.
2. Solve the active network for \(V_{oc}\); this is \(V_{Th}\).
3. For a network containing only independent sources, set independent voltage sources to zero by shorting them and independent current sources to zero by opening them; the resistance seen into the terminals is \(R_{Th}\).
4. Keep every dependent source active. Apply a test voltage \(V_t\) or test current \(I_t\) at the terminals and use \(R_{Th}=V_t/I_t\). Alternatively, when permitted, \(R_{Th}=V_{oc}/I_{sc}\).
5. Obtain \(I_N=V_{Th}/R_{Th}\) and redraw either equivalent before reconnecting \(R_L\).

Maximum-power theorem¶

For a resistive DC network, the load power is

Differentiating with respect to \(R_L\) and setting \(dP_L/dR_L=0\) gives

At this condition the load and source-equivalent resistances dissipate equal power, so efficiency is \(50\%\). For AC impedances, maximum average power requires conjugate matching, \(Z_L=Z_{Th}^{*}\).

Worked application: a \(12\,\text{V}\) ideal source feeds \(4\,\Omega\) in series to an output node, with \(6\,\Omega\) from that node to ground. With the load removed,

After shorting the independent \(12\,\text{V}\) source,

Thus \(I_N=7.2/2.4=3\,\text{A}\) and \(R_N=2.4\,\Omega\). For maximum power choose \(R_L=2.4\,\Omega\):

The load voltage is \(3.6\,\text{V}\) and \(P_L=(1.5)^2(2.4)=5.4\,\text{W}\), which checks the result. These equivalents are especially useful when one fixed source network must be evaluated for many loads; maximum-power matching is used in communication and electronic interfaces rather than efficient power distribution.

Practice target: 16–18 minutes; show source suppression, both equivalents, the power derivation, and one complete load calculation.

Exam-ready answer

A three-terminal resistive network may be connected as a star (Y), where \(R_1,R_2,R_3\) meet at a common neutral point, or as a delta (\(\Delta\)), where \(R_A,R_B,R_C\) form a closed triangle. Conversion preserves the resistance measured between every pair of external terminals and simplifies bridge networks that are not reducible by ordinary series-parallel rules.

Let \(S_\Delta=R_A+R_B+R_C\). Each star arm is the product of the two adjacent delta arms divided by their sum:

Let \(S_Y=R_1R_2+R_2R_3+R_3R_1\). Each delta arm is the sum of star pair-products divided by the opposite star arm:

Numerical check: for \(R_1=2\,\Omega\), \(R_2=3\,\Omega\), \(R_3=6\,\Omega\), \(S_Y=36\,\Omega^2\). Hence \(R_A=18\,\Omega\), \(R_B=12\,\Omega\), and \(R_C=6\,\Omega\). Converting back, \(S_\Delta=36\,\Omega\) and \(R_1=(12)(6)/36=2\,\Omega\), \(R_2=(18)(6)/36=3\,\Omega\), \(R_3=(18)(12)/36=6\,\Omega\), confirming terminal equivalence.

Therefore the essential memory rules are “adjacent product over delta sum” and “star pair-products over the opposite arm”; the formulas apply equally to impedances in a balanced-frequency AC network.

Practice target: 6–7 minutes; reproduce the labeled figure, both formula sets, and the balanced shortcut without notes.