DC Circuits Notes FE Sem 1 BEE | Mumbai University

Module 1 of Basic Electrical and Electronics Engineering deals with DC networks that contain only independent sources. It starts from Kirchhoff's laws and ideal versus practical sources, then builds the standard tools for solving any linear resistive network: source transformation, mesh and nodal analysis and star-delta conversion. Finally it covers the superposition, Thevenin, Norton and maximum power transfer theorems. This module carries the highest weightage of numericals in the paper and every later topic in AC circuits uses the same methods.

Summary

Kirchhoff's Current Law (KCL) states that the algebraic sum of currents meeting at a node is zero, i.e. total current entering a node equals total current leaving it; it follows from conservation of charge. Kirchhoff's Voltage Law (KVL) states that the algebraic sum of all voltage drops and EMFs around any closed loop is zero; it follows from conservation of energy. Sign convention matters: going round the loop, a rise in potential (crossing a source from - to +) is taken positive and a fall in potential (crossing a resistor in the direction of assumed current, or a source from + to -) negative, applied consistently.

An ideal voltage source maintains a constant terminal voltage whatever current is drawn, so its internal resistance is zero. A practical voltage source has a small internal resistance in series, so its terminal voltage falls as load current increases (V = E - I r). An ideal current source delivers a constant current whatever the load, so its internal resistance is infinite; a practical current source has a large internal resistance in parallel with it, so the load current falls slightly as load resistance rises.

Source transformation converts a practical voltage source (V in series with R) into an equivalent practical current source (I = V/R in parallel with the same R) and vice versa. The two are equivalent only as seen from the external terminals. The direction of the current source arrow points towards the positive terminal of the voltage source. Ideal sources cannot be transformed because R would be zero or infinite.

Mesh analysis applies KVL to every mesh (a loop with no other loop inside it) using assumed mesh currents, usually all clockwise. For a network with b branches and n nodes the number of independent mesh equations is b - n + 1. In the syllabus only independent voltage and current sources are used and supermesh cases are excluded. The equations can be written in matrix form [R][I] = [V], where diagonal terms are self-resistances and off-diagonal terms are negative mutual resistances, and solved by elimination or Cramer's rule.

Nodal analysis applies KCL at each node except a chosen reference (ground) node, with node voltages as unknowns. A network with n nodes needs n - 1 equations. Each branch current is written as (difference of node voltages)/resistance. It is preferred when the circuit has fewer nodes than meshes or has current sources. Supernode cases are excluded from the syllabus.

Star-delta and delta-star transformations replace three resistors connected in delta (between three terminals) by three resistors in star (from each terminal to a common point), or the reverse, without changing the resistance seen between any pair of terminals. They are used to simplify bridge-type networks that cannot be reduced by series-parallel combinations. For a balanced network, R(star) = R(delta)/3.

The superposition theorem states that in a linear bilateral network with several independent sources, the response (current or voltage) in any element equals the algebraic sum of the responses caused by each source acting alone, with all other ideal voltage sources replaced by short circuits and ideal current sources by open circuits (internal resistances retained). It does not apply directly to power because power is not linear in current.

Thevenin's theorem replaces any linear two-terminal network by a single voltage source Vth (the open-circuit voltage across the terminals) in series with Rth (the resistance seen from the terminals with all independent sources replaced by their internal resistances). Norton's theorem replaces it by a current source IN (the short-circuit current) in parallel with RN = Rth, and Vth = IN Rth. The maximum power transfer theorem states that a load receives maximum power from a network when RL = Rth, and this maximum power is Vth^2/(4 Rth) with an efficiency of only 50 percent.

Mind map

Procedure for Thevenin's theorem

Important definitions

Formulas

Key points to remember

• Syllabus restricts this module to independent sources only; dependent sources, supermesh and supernode are not asked.
• Mesh analysis is applicable only to planar circuits; nodal analysis works for planar and non-planar circuits.
• Number of mesh equations = b - n + 1; number of nodal equations = n - 1.
• In mesh analysis, the current through a branch shared by two meshes is the difference of the two mesh currents (when both are clockwise).
• A negative answer for an assumed current simply means the actual direction is opposite to the assumed one.
• While finding Rth or applying superposition, short ideal voltage sources and open ideal current sources; never remove the internal resistance.
• Superposition, Thevenin and Norton apply only to linear bilateral networks.
• Superposition cannot be used to find power directly; find the total current first and then compute I^2 R.
• At maximum power transfer, half the power is lost in Rth, so efficiency is only 50 percent; power systems are not operated at this point, but communication and electronic circuits are.
• Source transformation is valid only for practical sources and only as far as the external circuit is concerned.
• In delta-to-star, the denominator is the sum of the delta resistances; in star-to-delta, the numerator is the sum of the pairwise products of star resistances.

Solved examples

Example 1. Using mesh analysis find the current in the 4 ohm resistor. Mesh 1 contains a 10 V source (driving clockwise current), a 2 ohm resistor and the 4 ohm resistor. Mesh 2 contains the same 4 ohm resistor, a 3 ohm resistor and a 5 V source whose polarity opposes clockwise current in mesh 2.

Assume clockwise mesh currents I1 and I2. The 4 ohm resistor is common to both meshes.

Mesh 1 (KVL): 10 - 2I1 - 4(I1 - I2) = 0, so 6I1 - 4I2 = 10.

Mesh 2 (KVL): -4(I2 - I1) - 3I2 - 5 = 0, so -4I1 + 7I2 = -5.

Determinant D = (6)(7) - (-4)(-4) = 42 - 16 = 26.

I1 = [(10)(7) - (-4)(-5)]/26 = (70 - 20)/26 = 1.923 A.

I2 = [(6)(-5) - (-4)(10)]/26 = (-30 + 40)/26 = 0.385 A.

Current in 4 ohm = I1 - I2 = 1.923 - 0.385 = 1.538 A, flowing in the direction of I1 (downward through the common branch).

Example 2. A 12 V source with 2 ohm series resistance and a 6 V source with 6 ohm series resistance are both connected (positive terminals towards node A) to node A. A 4 ohm resistor is connected from node A to the reference node. Find the node voltage VA and all branch currents using nodal analysis.

Take the negative terminals of both sources as the reference node. Apply KCL at A, writing all currents as leaving A:

(VA - 12)/2 + VA/4 + (VA - 6)/6 = 0.

Multiply by 12: 6VA - 72 + 3VA + 2VA - 12 = 0, so 11VA = 84 and VA = 7.636 V.

Current from 12 V source into A = (12 - 7.636)/2 = 2.182 A.

Current in 4 ohm = 7.636/4 = 1.909 A (A to reference).

Current from A into the 6 V source = (7.636 - 6)/6 = 0.273 A (the 6 V source is being charged).

Check KCL: 2.182 = 1.909 + 0.273. Verified.

Example 3. For the circuit of the previous example, find the current in the 4 ohm resistor using the superposition theorem.

Case 1: 12 V source acting alone (6 V source replaced by a short circuit).

4 ohm parallel 6 ohm = 24/10 = 2.4 ohm. Total resistance = 2 + 2.4 = 4.4 ohm.

Supply current = 12/4.4 = 2.727 A. Voltage across the parallel part = 2.727 x 2.4 = 6.545 V.

Current in 4 ohm, I' = 6.545/4 = 1.636 A (downward).

Case 2: 6 V source acting alone (12 V source shorted).

2 ohm parallel 4 ohm = 8/6 = 1.333 ohm. Total = 6 + 1.333 = 7.333 ohm.

Supply current = 6/7.333 = 0.818 A. Voltage across the parallel part = 0.818 x 1.333 = 1.091 V.

Current in 4 ohm, I'' = 1.091/4 = 0.273 A (downward).

By superposition, I = I' + I'' = 1.636 + 0.273 = 1.909 A downward, which matches nodal analysis.

Example 4. For the same circuit, find the current in the 4 ohm resistor using Thevenin's theorem and Norton's theorem. Also find the value of load resistance that would receive maximum power and the value of that power.

Remove the 4 ohm load between A and the reference node.

Vth: the two sources now form a single loop. Loop current = (12 - 6)/(2 + 6) = 0.75 A flowing from the 12 V source towards the 6 V source.

Vth = 12 - 0.75 x 2 = 10.5 V (check: 6 + 0.75 x 6 = 10.5 V).

Rth: short both voltage sources; 2 ohm is in parallel with 6 ohm, Rth = 12/8 = 1.5 ohm.

Thevenin: IL = 10.5/(1.5 + 4) = 1.909 A.

Norton: short A to reference; IN = 12/2 + 6/6 = 6 + 1 = 7 A (also Vth/Rth = 10.5/1.5 = 7 A). RN = 1.5 ohm.

IL = 7 x 1.5/(1.5 + 4) = 1.909 A.

Maximum power transfer: RL = Rth = 1.5 ohm, Pmax = Vth^2/(4Rth) = 10.5^2/(4 x 1.5) = 110.25/6 = 18.375 W.

Example 5. Three resistors RAB = 10 ohm, RBC = 20 ohm and RCA = 30 ohm are connected in delta. Find the equivalent star network and verify by converting back.

Sum of delta resistances = 10 + 20 + 30 = 60 ohm.

RA = RAB x RCA/60 = 10 x 30/60 = 5 ohm.

RB = RAB x RBC/60 = 10 x 20/60 = 3.333 ohm.

RC = RBC x RCA/60 = 20 x 30/60 = 10 ohm.

Verification (star to delta): sum of pairwise products = 5 x 3.333 + 3.333 x 10 + 10 x 5 = 16.667 + 33.333 + 50 = 100.

RAB = 100/RC = 100/10 = 10 ohm, RBC = 100/RA = 100/5 = 20 ohm, RCA = 100/RB = 100/3.333 = 30 ohm. The original delta is recovered.

Example 6. A 10 V source with 5 ohm series resistance is connected in parallel with a 3 A current source having 10 ohm parallel resistance (both driving current towards the same terminal). Using source transformation, reduce the network to a single voltage source and find the current in a 6.667 ohm load connected across it.

Convert the voltage source: I = 10/5 = 2 A in parallel with 5 ohm.

Now two current sources in parallel aiding: 2 + 3 = 5 A. Parallel resistance = 5 x 10/(5 + 10) = 3.333 ohm.

Convert back to a voltage source: V = 5 x 3.333 = 16.667 V in series with 3.333 ohm.

Load current = 16.667/(3.333 + 6.667) = 16.667/10 = 1.667 A.

Example 7. A battery of EMF 12 V and internal resistance 0.2 ohm supplies a load of 5.8 ohm. Find the current, the terminal voltage and the equivalent practical current source.

Current I = E/(r + RL) = 12/(0.2 + 5.8) = 2 A.

Terminal voltage Vt = E - I r = 12 - 2 x 0.2 = 11.6 V (also I RL = 2 x 5.8 = 11.6 V).

Equivalent current source: Is = E/r = 12/0.2 = 60 A in parallel with 0.2 ohm.

Check: load current = 60 x 0.2/(0.2 + 5.8) = 2 A, the same as before.

Important questions and answers

Q1. State and explain Kirchhoff's laws. (4 marks)

KCL: the algebraic sum of currents meeting at a node is zero; total current entering equals total current leaving. Example: if I1 and I2 enter a node and I3 leaves, I1 + I2 = I3. It is based on conservation of charge since charge cannot accumulate at a node.

KVL: in any closed loop the algebraic sum of EMFs equals the algebraic sum of IR drops, i.e. the sum of all voltages around the loop is zero. It is based on conservation of energy.

Sign convention: going around the loop, a rise in potential (from - to + of a source) is positive and a fall (across a resistor in the direction of current, or from + to - of a source) is negative. Both laws apply to linear and non-linear, DC and AC circuits.

Q2. Differentiate between ideal and practical voltage and current sources. (4 marks)

Ideal voltage source: zero internal resistance; terminal voltage constant for all load currents; V-I characteristic is a horizontal line.

Practical voltage source: small internal resistance r in series; terminal voltage V = E - I r falls with load current; characteristic slopes downward.

Ideal current source: infinite internal resistance; supplies constant current for any load voltage; characteristic is a vertical line.

Practical current source: large internal resistance in parallel; load current decreases slightly as load voltage increases.

A practical voltage source is close to ideal when r is much smaller than RL; a practical current source is close to ideal when its parallel resistance is much larger than RL.

Q3. Explain source transformation with a neat diagram. (3 marks)

A practical voltage source V with series resistance R can be replaced by a current source I = V/R with the same R connected in parallel, and vice versa, because both give the same terminal voltage and current to any external load. The arrow of the current source points towards the + terminal of the original voltage source. The transformation is not valid for ideal sources and the internal currents and powers of the two equivalents are not the same; the equivalence holds only for the external circuit. It is used to combine sources and simplify networks before applying mesh or nodal analysis.

Q4. Derive the expressions for converting a delta network into an equivalent star network. (6 marks)

Let delta resistances RAB, RBC, RCA be connected between terminals A, B, C and the equivalent star have arms RA, RB, RC to a common point N. For equivalence, the resistance between any two terminals with the third open must be equal.

Between A and B: RA + RB = RAB(RBC + RCA)/(RAB + RBC + RCA) ...(1)

Between B and C: RB + RC = RBC(RCA + RAB)/(RAB + RBC + RCA) ...(2)

Between C and A: RC + RA = RCA(RAB + RBC)/(RAB + RBC + RCA) ...(3)

Adding (1), (2), (3) and dividing by 2: RA + RB + RC = (RAB RBC + RBC RCA + RCA RAB)/(RAB + RBC + RCA) ...(4)

Subtract (2) from (4): RA = RAB RCA/(RAB + RBC + RCA).

Subtract (3) from (4): RB = RAB RBC/(RAB + RBC + RCA).

Subtract (1) from (4): RC = RBC RCA/(RAB + RBC + RCA).

Rule: each star arm equals the product of the two delta resistances meeting at that terminal divided by the sum of the three delta resistances. For a balanced delta, RY = RD/3.

Q5. State and explain the superposition theorem. What are its limitations? (5 marks)

Statement: In any linear bilateral network containing two or more independent sources, the current through (or voltage across) any element is the algebraic sum of the currents (or voltages) produced by each source acting alone, all other independent sources being replaced by their internal resistances.

Procedure: (1) keep one source, short-circuit other ideal voltage sources and open-circuit ideal current sources; (2) find the required current due to this source; (3) repeat for every source; (4) add the individual currents algebraically, taking directions into account.

Limitations: applicable only to linear bilateral networks; cannot be used to calculate power directly because power is proportional to the square of current; not applicable to networks with non-linear elements such as diodes; it becomes lengthy when there are many sources.

Q6. State and explain Thevenin's theorem with steps. (5 marks)

Statement: Any linear bilateral two-terminal network containing sources and resistances can be replaced by an equivalent circuit consisting of a voltage source Vth in series with a resistance Rth, where Vth is the open-circuit voltage at the terminals and Rth is the resistance seen from the terminals when all independent sources are replaced by their internal resistances.

Steps: (1) remove the load RL and mark terminals A-B; (2) find the open-circuit voltage VAB = Vth; (3) replace sources by internal resistances (voltage sources shorted, current sources opened) and find Rth looking into A-B; (4) draw the Thevenin equivalent and reconnect RL; (5) IL = Vth/(Rth + RL).

Advantage: when RL changes, only the last step needs repeating, so it is ideal for finding current in a variable load.

Q7. State and explain Norton's theorem. How are Thevenin and Norton equivalents related? (5 marks)

Statement: Any linear bilateral two-terminal network can be replaced by an equivalent current source IN in parallel with a resistance RN, where IN is the current through a short circuit placed across the terminals and RN is the resistance seen from the terminals with all independent sources replaced by their internal resistances.

Steps: (1) remove RL and short-circuit A-B; (2) find the short-circuit current IN; (3) find RN exactly as Rth; (4) draw IN parallel RN and reconnect RL; (5) IL = IN RN/(RN + RL).

Relation: RN = Rth and Vth = IN Rth, so one equivalent can be obtained from the other by source transformation.

Q8. State and prove the maximum power transfer theorem. (6 marks)

Statement: A resistive load receives maximum power from a linear DC network when the load resistance equals the Thevenin resistance of the network as seen from the load terminals.

Proof: Replace the network by Vth in series with Rth. Load current I = Vth/(Rth + RL). Power in load P = I^2 RL = Vth^2 RL/(Rth + RL)^2.

For maximum P, dP/dRL = 0: Vth^2[(Rth + RL)^2 - 2RL(Rth + RL)]/(Rth + RL)^4 = 0, so (Rth + RL) - 2RL = 0, giving RL = Rth. (The second derivative is negative, so this is a maximum.)

Maximum power: Pmax = Vth^2 Rth/(2Rth)^2 = Vth^2/(4Rth).

Efficiency at maximum power = power in RL/total power = I^2 RL/(I^2 (RL + Rth)) = 50 percent. Hence it is used in communication and electronic circuits (matching speakers, antennas) but not in power systems where high efficiency is essential.

Q9. Compare mesh analysis and nodal analysis. (4 marks)

Mesh analysis is based on KVL; nodal analysis on KCL.

Unknowns: mesh currents versus node voltages.

Number of equations: b - n + 1 for mesh, n - 1 for nodal.

Mesh analysis applies only to planar networks; nodal analysis applies to planar and non-planar networks.

Mesh is convenient when the circuit has voltage sources and fewer meshes; nodal is convenient when there are current sources, many parallel branches and fewer nodes.

Both give the same branch currents and voltages for a given network.

Q10. Explain the procedure for mesh analysis and write the equations in matrix form. (4 marks)

(1) Identify the meshes and assign a clockwise current to each. (2) Apply KVL to each mesh; the current in a shared resistor is the difference of the two mesh currents. (3) Arrange the equations in the form [R][I] = [V]. For two meshes: [R11 -R12; -R21 R22][I1; I2] = [V1; V2], where R11, R22 are self-resistances (sum of resistances in the mesh), R12 = R21 is the mutual (shared) resistance, and V1, V2 are the net source voltages driving current in the clockwise direction in each mesh. (4) Solve by Cramer's rule or elimination. (5) Find branch currents from the mesh currents; a negative value means the actual direction is opposite.

Exam tips

• Always draw the circuit and mark assumed current directions and polarities before writing equations; examiners award marks for a neat circuit at each step.
• For Thevenin and Norton numericals, draw three separate diagrams: circuit for Vth (or IN), circuit for Rth, and the final equivalent with the load. Each carries marks.
• Write mesh equations in standard form and use Cramer's rule; show the determinant values.
• In superposition, label each partial current with its direction (I', I'') before adding algebraically.
• Check your final answer with KCL at a node or by solving the same circuit with a second method if time permits.
• Proof of maximum power transfer and derivation of star-delta are frequent 5-6 mark theory questions; memorise the derivation steps.
• Units matter: write ohm, A, V and W with every numerical answer.

FAQs

When should I use mesh analysis and when nodal analysis?

Count the equations: mesh needs b - n + 1 and nodal needs n - 1. Choose the method with fewer equations. Circuits with current sources usually favour nodal analysis, while series-type circuits with voltage sources favour mesh analysis.

Why is the efficiency only 50 percent at maximum power transfer?

Because RL = Rth, the same current flows through equal resistances, so the power lost inside the source (I^2 Rth) equals the power delivered to the load (I^2 RL).

What do I do with a current source while finding Rth?

Replace an ideal current source by an open circuit (its internal resistance is infinite) and an ideal voltage source by a short circuit. If a practical source is given, keep its internal resistance in the circuit.

Are dependent sources, supermesh or supernode asked in this module?

No. The MU syllabus explicitly limits this module to independent sources and excludes supermesh and supernode cases.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Basic Electrical & Electronics Engineering) and were prepared with AI help and checked.