Basic Electrical & Electronics Engineering (BT-104) Notes

How unit 1 is examined

This unit covers sources, Ohm's and Kirchhoff's laws, the network theorems and the analysis methods for D.C. circuits; mesh and nodal analysis, superposition and star-delta carry the most marks, with Thevenin, Kirchhoff and sources next.

Voltage and current sources

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Not asked since 2022</span>

Definition. A source is an active element that supplies energy to a circuit; a voltage source keeps its terminal voltage fixed and a current source keeps its branch current fixed.

Key points.

An ideal voltage source has zero internal resistance, so its voltage is independent of the load current.
A practical voltage source is an ideal source $E$ in series with an internal resistance $R_s$, so its terminal voltage is $V = E - IR_s$.
An ideal current source has infinite internal resistance and delivers the same current whatever the load.
A practical current source is an ideal source $I_s$ in parallel with a resistance $R_p$.

Dependent and independent sources

Definition. An independent source gives a voltage or current that does not depend on any other quantity in the circuit; a dependent (controlled) source gives a value fixed by a voltage or current elsewhere in the circuit.

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Key points.

An independent source is drawn as a circle: a voltage source with the + and - marks, a current source with an arrow.
Independent sources may be ideal (no internal resistance for voltage, infinite for current) or practical (with $R_s$ in series or $R_p$ in parallel).
A dependent source is drawn as a diamond and is used to model transistors and amplifiers.
There are four dependent types, listed in the table below.
A dependent source cannot be switched off (deactivated) in superposition or Thevenin's theorem; it stays in the circuit.

Type	Full name	Output
VCVS	voltage controlled voltage source	$v = \mu v_x$
CCVS	current controlled voltage source	$v = r i_x$
VCCS	voltage controlled current source	$i = g v_x$
CCCS	current controlled current source	$i = \beta i_x$

Answer frame. Open with the definition of a source; draw ideal and practical voltage and current sources, then the four diamond symbols; develop points 1-5 with the table; close with the voltage-to-current conversion (see Source conversion).

Asked: [7 marks] (Jun 2023, Dec 2024) What do you understand by Source. Discuss different types of dependent and independent voltage and current sources with suitable sketch. Also, how can a voltage source be converted into a current source?

Units and dimensions

Definition. Electrical quantities are measured in SI units, each expressed through the base dimensions mass $M$, length $L$, time $T$ and current $A$.

Key points.

Charge is in coulomb (C), current in ampere (A = C/s), voltage in volt (V = J/C) and resistance in ohm ($\Omega$ = V/A).
Power is in watt (W = J/s) and energy in joule (J); one kilowatt-hour equals $3.6\times10^{6}$ J.
Dimensions of voltage are $[ML^2T^{-3}A^{-1}]$ and of resistance $[ML^2T^{-3}A^{-2}]$.

Source conversion

Definition. A practical voltage source $E$ with series resistance $R_s$ is equivalent, at its terminals, to a current source $I_s = E/R_s$ with the same resistance $R_s$ in parallel.

Key points.

To go from voltage to current source, set $I_s = E/R_s$ and place $R_s$ in parallel; the reverse gives $E = I_s R_s$ with $R_s$ in series.
The arrow of the current source points toward the + terminal of the voltage source.
The two are equivalent only for the external load; the internal power loss differs.
An ideal voltage source (R = 0) cannot be converted.

Ohm’s Law

Definition. ==At constant temperature the current through a conductor is directly proportional to the potential difference across it, $V = IR$.==

Key points.

The constant of proportionality $R$ is the resistance, measured in ohms.
Ohm's law holds only for linear (ohmic) elements at constant temperature; diodes and lamps do not obey it.
The equivalent forms are $I = V/R$ and $R = V/I$.
Resistance is $R = \rho l/a$: it rises with length and resistivity and falls with area.

Kirchhoff’s Law

Definition. KCL: the algebraic sum of the currents meeting at a node is zero. KVL: the algebraic sum of the voltages around any closed loop is zero.

Diagram. <figure class="ds-fig" style="margin:1.4rem 0;overflow-x:auto"><svg xmlns="http://www.w3.org/2000/svg" id="dsfig-u1-02" viewBox="0 0 424 252" width="424" height="252" role="img" aria-label="KCL at node N: I1 enters, I2 and I3 leave, so I1 = I2 + I3"><style>#dsfig-u1-02 .e{stroke:#454C5A;stroke-width:1.4;fill:none}#dsfig-u1-02 .e.hi{stroke:#2340B8;stroke-width:2.6}#dsfig-u1-02 .n{fill:#FFFFFF;stroke:#16181D;stroke-width:1.4}#dsfig-u1-02 .n.hi{fill:#E3E9FC;stroke:#2340B8;stroke-width:2.2}#dsfig-u1-02 .n.rb-b{fill:#16181D;stroke:#16181D}#dsfig-u1-02 .n.rb-r{fill:#BD3227;stroke:#BD3227}#dsfig-u1-02 text{font-family:"JetBrains Mono",ui-monospace,Menlo,Consolas,monospace;font-size:13px}#dsfig-u1-02 .t{fill:#16181D;font-weight:500}#dsfig-u1-02 .t.inv{fill:#FFFFFF;font-weight:700}#dsfig-u1-02 .kd{stroke:#16181D;stroke-width:1.2}#dsfig-u1-02 .dot{fill:#16181D}#dsfig-u1-02 .ann{fill:#2340B8;font-size:11px;font-weight:700}#dsfig-u1-02 .lbl{fill:#6F7787;font-family:system-ui,-apple-system,sans-serif;font-size:12px;font-weight:700}#dsfig-u1-02 .ptr{fill:#2340B8;font-size:12px;font-weight:700}#dsfig-u1-02 .ah{fill:#454C5A}#dsfig-u1-02 .ah.hi{fill:#2340B8}#dsfig-u1-02 .wl rect{fill:#FFFFFF;stroke:#DCE0E7}#dsfig-u1-02 .wl .t{font-size:12px;font-weight:700}#dsfig-u1-02 .wl.hi rect{fill:#2340B8;stroke:#2340B8}#dsfig-u1-02 .wl.hi .t{fill:#FFFFFF}html.dark #dsfig-u1-02 .e{stroke:#B1B7C3}html.dark #dsfig-u1-02 .e.hi{stroke:#8FA3FF}html.dark #dsfig-u1-02 .n{fill:#161920;stroke:#E6E8ED}html.dark #dsfig-u1-02 .n.hi{fill:#1E2748;stroke:#8FA3FF}html.dark #dsfig-u1-02 .n.rb-b{fill:#E6E8ED;stroke:#E6E8ED}html.dark #dsfig-u1-02 .n.rb-r{fill:#FF7E71;stroke:#FF7E71}html.dark #dsfig-u1-02 .t{fill:#E6E8ED}html.dark #dsfig-u1-02 .t.inv{fill:#0F1115}html.dark #dsfig-u1-02 .kd{stroke:#E6E8ED}html.dark #dsfig-u1-02 .dot{fill:#E6E8ED}html.dark #dsfig-u1-02 .ann{fill:#8FA3FF}html.dark #dsfig-u1-02 .lbl{fill:#858D9C}html.dark #dsfig-u1-02 .ptr{fill:#8FA3FF}html.dark #dsfig-u1-02 .ah{fill:#B1B7C3}html.dark #dsfig-u1-02 .ah.hi{fill:#8FA3FF}html.dark #dsfig-u1-02 .wl rect{fill:#161920;stroke:#2A2E37}html.dark #dsfig-u1-02 .wl.hi rect{fill:#8FA3FF;stroke:#8FA3FF}html.dark #dsfig-u1-02 .wl.hi .t{fill:#0F1115}</style><defs><marker id="ah2" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah" d="M0,1 L9,5 L0,9 z"/></marker><marker id="ahh2" viewBox="0 0 10 10" refX="9" refY="5" markerWidth="7" markerHeight="7" orient="auto-start-reverse"><path class="ah hi" d="M0,1 L9,5 L0,9 z"/></marker></defs><path class="e" d="M59,126 L191,126" marker-end="url(#ah2)"/><path class="e" d="M229,117.5 L365.2,49.4" marker-end="url(#ah2)"/><path class="e" d="M229,134.5 L365.2,202.6" marker-end="url(#ah2)"/><circle class="n" cx="212" cy="126" r="18"/><text class="t" x="212" y="126" dy=".35em" text-anchor="middle">N</text><circle class="n" cx="40" cy="126" r="18"/><text class="t" x="40" y="126" dy=".35em" text-anchor="middle">A</text><circle class="n" cx="384" cy="40" r="18"/><text class="t" x="384" y="40" dy=".35em" text-anchor="middle">B</text><circle class="n" cx="384" cy="212" r="18"/><text class="t" x="384" y="212" dy=".35em" text-anchor="middle">C</text></svg><figcaption style="font-size:.82em;opacity:.72;margin-top:.45rem">KCL at node N: I1 enters, I2 and I3 leave, so I1 = I2 + I3</figcaption></figure>

Key points.

KCL follows from conservation of charge: charge cannot accumulate at a node, so $\sum I_{in} = \sum I_{out}$.
Currents entering a node are taken positive and leaving negative, so $\sum I = 0$.
KVL follows from conservation of energy: a charge returning to its start has gained and lost equal energy.
Sign rule for KVL: a rise (moving from - to + of a source) is positive; a drop across a resistor in the direction of assumed current is negative, giving $\sum E = \sum IR$.
Example: in a series loop of a 12 V source with 2 $\Omega$ and 4 $\Omega$, KVL gives $12 - 2I - 4I = 0$, so $I = 2$ A.
Both laws hold for linear and non-linear, and for D.C. and A.C., circuits.

Answer frame. Open with the two statements; draw a node with three branch currents and a one-source two-resistor loop; write KCL with sign convention, then KVL with sign rule, then the loop example; close by saying these two laws are the basis of mesh and nodal analysis.

Asked: [7 marks] (Jun 2022, Jun 2025) State and explain KCL and KVL with suitable example (Kirchhoff's current and voltage law).

Superposition theorem

Definition. <mark>In a linear bilateral network with more than one independent source, the response (current or voltage) in any element equals the algebraic sum of the responses produced by each independent source acting alone.</mark>

Key points.

It applies only to linear networks; power is not linear ($P = I^2R$), so power is found after the currents are added.
Take one independent source at a time and deactivate the others.
A voltage source is deactivated by replacing it with a short circuit, a current source by an open circuit; internal resistances stay.
Dependent sources are never deactivated; they stay in every case.
Add the individual responses algebraically, taking each direction into account.
It needs one circuit solved per source, so it suits circuits with two or three sources.

Steps. (1) Keep source 1, deactivate the rest, find the response $I'$. (2) Repeat for each other source to get $I''$. (3) Add: $I = I' + I''$. (4) Use the total in the power or voltage formula.

Example. Power in the 9 $\Omega$ resistor (32 V source with 4 $\Omega$, 12 $\Omega$ branch, 6 $\Omega$ link, 4 A source, 9 $\Omega$). Take downward current in 9 $\Omega$ as positive.

Case	Working	Result
32 V alone (4 A open)	$V_{9} = -12$ V, from node equations	$I' = 4/3$ A upward
4 A alone (32 V shorted)	$V_{9} = 18$ V	$I'' = 2$ A downward
Both	$I = -4/3 + 2$	$I = 2/3$ A

$$P = I^2 R = \left(\tfrac{2}{3}\right)^2 \times 9$$

P = 4 W.

Answer frame. For the explain question: open with the statement, give the four conditions and the deactivation rule, then the steps, and close with the note that power is not superposed. For the numerical: draw the original circuit and one redrawn circuit per source, solve each, add with signs, then find the required power or voltage.

Pitfall: Adding powers of the individual cases instead of adding currents first loses the marks.

Asked: [7 marks] (Jun 2022, Dec 2023, Dec 2024) Compute the power dissipated in the 9 $\Omega$ resistor using superposition theorem; using superposition find $V_0$ in the given circuit; find the current in branch AB by superposition. Asked: [7 marks] (Jun 2025) State and explain Superposition theorem.

Thevenin’s theorem and their application for analysis of series and parallel resistive circuits excited by independent voltage sources

Definition. <mark>Any linear two-terminal network can be replaced by a single voltage source $V_{th}$ in series with a resistance $R_{th}$, where $V_{th}$ is the open-circuit voltage across the terminals and $R_{th}$ is the resistance seen from them with all independent sources deactivated.</mark>

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Key points.

Remove the load resistor and mark the two terminals $a$, $b$.
Find $V_{th}$ as the open-circuit voltage between $a$ and $b$ using KVL, mesh or divider rules.
Find $R_{th}$ by shorting voltage sources and opening current sources, then reducing series and parallel branches as seen from $a$-$b$.
If a dependent source is present, do not deactivate it: apply a test source, $R_{th} = V/I$, or use $R_{th} = V_{th}/I_{sc}$.
Redraw $V_{th}$ and $R_{th}$ with the load reconnected and find $I_L = V_{th}/(R_{th} + R_L)$.
Its advantage is that the load can change many times while the rest of the network is solved once.

Example. A 6 V source, 2 $\Omega$ load $R_1$ in series with a branch of 4 $\Omega$ parallel to a 3 A current source. Open-circuit: the right node sits at $3\times4 = 12$ V, the left at 6 V, so $V_{th} = 6$ V. Deactivate: $R_{th} = 4\ \Omega$.

$$I = \frac{V_{th}}{R_{th}+R_1} = \frac{6}{4+2}$$

I = 1 A.

Answer frame. Open with the statement; draw the original circuit, the circuit with load removed, and the Thevenin equivalent with load; do the steps in order, $V_{th}$ then $R_{th}$ then $I_L$; close with the final current and its direction.

Asked: [7 marks] (Nov 2022, Jun 2023) Find the current through the 6 $\Omega$ resistor using Thevenin's theorem (circuit has a dependent source $2V_x$); find the current through $R_1$ using Thevenin's theorem.

Power & Energy in such circuits

Definition. Power is the rate of doing electrical work, $P = VI$ watts; energy is power multiplied by time, $W = Pt$ joules.

Key points.

In a resistor, $P = VI = I^2R = V^2/R$, and all of it is dissipated as heat.
A source delivering current $I$ at voltage $V$ supplies $P = VI$; a source being charged absorbs it.
In any circuit, total power supplied equals total power dissipated.
Energy in kWh is $P(\text{kW}) \times t(\text{h})$; one kWh is one unit on the bill.

Mesh & nodal analysis

Definition. <mark>Mesh analysis applies KVL to the independent meshes to solve the mesh currents; nodal analysis applies KCL at the independent nodes to solve the node voltages.</mark>

Key points.

A mesh is a loop with no other loop inside it; mesh analysis needs one KVL equation per mesh.
Assume all mesh currents clockwise. The self-resistance of a mesh is the sum of its resistors, with positive sign, and each shared resistor enters with the other mesh current with negative sign.
The equations form the matrix $[R][I] = [V]$, solved by Cramer's rule or substitution; a branch current is the difference of the two mesh currents through it.
If a current source lies in one mesh, that mesh current is fixed; if it is shared, combine the two meshes into a supermesh and add the constraint equation.
Nodal analysis: choose a reference node, assign voltages to the others, and write KCL as (sum of currents leaving) = 0 with each current $(V_n - V_m)/R$.
A branch with a voltage source and resistor is written as $(V_n - E)/R$; a dependent source is expressed in terms of node voltages or mesh currents and substituted.
Choose mesh analysis when meshes are fewer than nodes and nodal when nodes are fewer.

Example. Circuit of Jun 2025 Fig (i): 35 V source, 12 $\Omega$ from B to C, 2 $\Omega$ from C to the rail, 4 $\Omega$ from C to D, 40 V source at D.

Nodal, reference at the bottom rail:

$$\frac{V_C-35}{12}+\frac{V_C}{2}+\frac{V_C-40}{4}=0 \Rightarrow 10V_C = 155$$

$V_C = 15.5$ V, so $I_{2\Omega} = 15.5/2 = 7.75$ A.

Mesh check (clockwise): $14I_1 - 2I_2 = 35$ and $-2I_1 + 6I_2 = -40$ give $I_1 = 1.625$ A, $I_2 = -6.125$ A, and the 2 $\Omega$ current is $I_1 - I_2 = 7.75$ A.

Answer frame. For a mesh numerical: draw the circuit with clockwise mesh currents labelled, write one KVL per mesh, put them in $[R][I]=[V]$, solve by Cramer's rule, and state each current with unit and direction. For a nodal numerical: mark the reference node, write KCL at each unknown node in $(V-E)/R$ form, solve, then find the asked current from a node-voltage difference. Close with the boxed answer.

Pitfall: Forgetting the negative sign on the shared resistor term, or the sign of a source that opposes the mesh current, gives a wrong matrix.

Asked: [7 marks] (Dec 2023, Jun 2023, Jun 2025) Use mesh analysis to determine the three mesh currents; find $I_0$ by mesh analysis (dependent source $10 i_0$, supermesh); find the current through the 5 $\Omega$ resistance using mesh current analysis. Asked: [7 marks] (Nov 2022) Find the voltage $V_{ab}$ in the network shown (30 V source with 4 $\Omega$ and 6 $\Omega$, 2 A source, 10 $\Omega$ resistor). Asked: [7 marks] (Dec 2024) Using nodal analysis, find the current through the 10 $\Omega$ resistor (15 V and 30 V sources). Asked: [7 marks] (Jun 2025) For the circuit in Fig (i) find the current in the 2 $\Omega$ resistor.

Star Delta transformation & circuits

Definition. <mark>A star (Y) network of three resistors joined at a common point can be replaced by an equivalent delta ($\Delta$) of three resistors joined in a closed triangle, and vice versa, so that the resistance between any two terminals is the same.</mark>

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Key points.

It is used when resistors are neither in series nor in parallel, as in a bridge, and simplifies them into series-parallel form.
The equivalence is set by equating the resistance between each pair of terminals in the two networks.
Delta to star: each star resistor is the product of the two adjacent delta resistors divided by the sum of all three.
Star to delta: each delta resistor is the sum of the two adjacent star resistors plus their product divided by the third star resistor.
If all three resistors are equal, $R_Y = R_\Delta/3$ and $R_\Delta = 3R_Y$.
After the reduction, the rest of the network is found by ordinary series and parallel combination.

Formula. $$R_1=\frac{R_bR_c}{R_a+R_b+R_c},\quad R_2=\frac{R_cR_a}{R_a+R_b+R_c},\quad R_3=\frac{R_aR_b}{R_a+R_b+R_c}$$

$$R_a=R_1+R_2+\frac{R_1R_2}{R_3},\quad R_b=R_2+R_3+\frac{R_2R_3}{R_1},\quad R_c=R_3+R_1+\frac{R_3R_1}{R_2}$$

Example. A delta of three 30 $\Omega$ resistors: $R_Y = \dfrac{30\times30}{90} = 10\ \Omega$ each. Between two terminals the star gives $10+10 = 20\ \Omega$, and the delta gives $30 \parallel 60 = 20\ \Omega$, so both agree.

Answer frame. Open with the definition of star and delta; draw both networks with matching terminals labelled; write the two sets of formulas; give the equal-resistance case and the 30 $\Omega$ example; close by naming the use, bridge reduction.

Asked: [14 marks] (Dec 2023, Dec 2024) Discuss the star-delta transformation using suitable example; write a short note on star-delta transformation (as one of two).

Last-minute revision

Ohm's law: $V = IR$; $P = VI = I^2R = V^2/R$; $W = Pt$.
KCL: $\sum I = 0$ at a node; KVL: $\sum V = 0$ around a loop.
Voltage source to current source: $I_s = E/R_s$, $R_s$ in parallel.
Four dependent sources: VCVS, CCVS, VCCS, CCCS.
Superposition: linear only; short voltage sources, open current sources; add currents, not powers.
Superposition paper answer: $I' = 4/3$ A, $I'' = 2$ A, $I = 2/3$ A, $P = 4$ W.
Thevenin: $I_L = V_{th}/(R_{th}+R_L)$; dependent sources are not deactivated.
Mesh: $[R][I] = [V]$, clockwise currents; current source in a shared branch means supermesh.
Nodal: KCL at each unknown node; Jun 2025 gives $V_C = 15.5$ V and $I = 7.75$ A.
Delta to star: $R_1 = R_bR_c/(R_a+R_b+R_c)$; equal resistors $R_Y = R_\Delta/3$.
1 kWh = 1 unit = $3.6\times10^6$ J.

Memory hooks

Star from delta: "product over sum" (two adjacent over total). Delta from star: "sum plus product over the opposite".
Superposition: "Short the Volts, Open the Amps."
KCL is charge (nodes), KVL is energy (loops).
Mesh = KVL with currents; Nodal = KCL with voltages.
Thevenin: open for $V$, dead sources for $R$.

Coverage checklist

Voltage and current sources: definition, ideal and practical.
dependent and independent sources: Jun 2023, Dec 2024 (7 marks).
Units and dimensions: SI units and dimensions.
Source Conversion: voltage to current, part of the Jun 2023 and Dec 2024 question.
Ohm’s Law: statement and limits.
Kirchhoff’s Law: Jun 2022, Jun 2025 (7 marks).
Superposition theorem: Jun 2022, Dec 2023, Dec 2024, Jun 2025 (7 marks each).
Thevenin's theorem and their application for analysis of series and parallel resistive circuits excited by independent voltage sources: Nov 2022, Jun 2023 (7 marks).
Power & Energy in such circuits: formulas and kWh.
Mesh & nodal analysis: Nov 2022, Jun 2023, Dec 2023, Dec 2024, Jun 2025 (7 marks each).
Star Delta transformation & circuits: Dec 2023, Dec 2024 (7 and 14 marks).

Account

Basic Electrical & Electronics Engineering (BT-104) - Unit 1 Short Notes

How unit 1 is examined

Voltage and current sources

Dependent and independent sources

Units and dimensions

Source conversion

Ohm’s Law

Kirchhoff’s Law

Superposition theorem

Thevenin’s theorem and their application for analysis of series and parallel resistive circuits excited by independent voltage sources

Power & Energy in such circuits

Mesh & nodal analysis

Star Delta transformation & circuits

Last-minute revision

Memory hooks

Coverage checklist