Unit 1: Fundamentals of D.C. circuits - Subjective Questions

ECE131 — Basic Electrical And Electronics Engineering • Practice Questions with Detailed Answers

20 questions

Define resistance, inductance, and capacitance. Explain the voltage-current relationship and steady-state DC behavior of each element.

Explain the concepts of voltage, current, electrical power, and electrical energy. A device operates at $12\,V$ and draws $2\,A$ for $3$ minutes. Calculate the power and energy consumed.

State and explain Ohm's law. Mention its limitations and calculate the current through a $10\,\Omega$ resistor connected across a $50\,V$ supply.

State Kirchhoff's Current Law and Kirchhoff's Voltage Law. Explain their physical basis and application in circuit analysis.

Describe the basic systematic method and the intuitive method of DC circuit analysis. Compare their advantages and limitations.

A $4\,\Omega$ resistor is connected in series with a parallel combination of $6\,\Omega$ and $3\,\Omega$ . The network is connected to a $12\,V$ source. Find the equivalent resistance, source current, and current through each parallel branch.

Derive the voltage division rule. Three resistors of $2\,\Omega$ , $3\,\Omega$ , and $5\,\Omega$ are connected in series across a $20\,V$ supply. Find the voltage across each resistor.

Derive the current division rule for two parallel resistors. A total current of $6\,A$ enters a parallel combination of $6\,\Omega$ and $3\,\Omega$ . Determine the branch currents.

Explain star-delta and delta-star transformations. Derive the conversion formulas and convert a balanced delta having $3\,\Omega$ in each branch into an equivalent star.

Distinguish between independent and dependent sources. Describe the four types of dependent sources used in electrical circuits.

Explain mesh analysis and solve a two-mesh circuit whose mesh equations are $6I_1-2I_2=10$ and $-2I_1+8I_2=5$ . Also find the current through the common branch.

Explain nodal analysis. A node $A$ is connected to a $12\,V$ node through $3\,\Omega$ , to ground through $6\,\Omega$ , and to node $B$ through $2\,\Omega$ . Node $B$ is connected to ground through $4\,\Omega$ . Find $V_A$ and $V_B$ .

Compare mesh analysis and nodal analysis. State the situations in which each method is preferable.

State Thevenin's theorem and explain the procedure for finding a Thevenin equivalent. A $12\,V$ source feeds a $4\,\Omega$ series resistor and an $8\,\Omega$ resistor connected to ground. Find the Thevenin equivalent across the $8\,\Omega$ resistor and determine the current when a $4\,\Omega$ load is connected across the terminals.

State Norton's theorem and explain its relationship with Thevenin's theorem. Convert a Thevenin source of $8\,V$ in series with $8/3\,\Omega$ into its Norton equivalent and find the current through a $4\,\Omega$ load.

State and derive the maximum power transfer theorem for a DC resistive circuit. Also determine the maximum power delivered by a Thevenin source having $V_{Th}=12\,V$ and $R_{Th}=3\,\Omega$ .

State the superposition theorem and explain its application. A node is connected to a $12\,V$ source through $6\,\Omega$ , to a $6\,V$ source through $3\,\Omega$ , and to ground through $2\,\Omega$ . Use superposition to find the node voltage.

Explain source transformation. Convert a $12\,V$ voltage source in series with $4\,\Omega$ into an equivalent current source, and show that both forms produce the same current in an $8\,\Omega$ load.

Derive the expressions for energy stored in an inductor and a capacitor. Calculate the energy stored in a $2\,H$ inductor carrying $3\,A$ and a $100\,\mu F$ capacitor charged to $20\,V$ .

Explain how a circuit containing a dependent source is analyzed. A node of voltage $v$ is connected to ground through a $5\,\Omega$ resistor. An independent $2\,A$ source and a dependent current source of value $0.1v\,A$ both inject current into the node. Find the node voltage and the power absorbed by the resistor.

Question 1

Define resistance, inductance, and capacitance. Explain the voltage-current relationship and steady-state DC behavior of each element.

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Answer 1

Resistance: Resistance is the opposition offered by a material to the flow of electric current. Its SI unit is the ohm $(\Omega)$ . For a linear resistor,

$v(t)=Ri(t)$

A resistor dissipates electrical energy as heat.

Inductance: Inductance is the property of a coil that opposes a change in current by inducing an emf. Its SI unit is the henry $(H)$ . Its voltage-current relation is

$v_L(t)=L\frac{di(t)}{dt}$

An ideal inductor behaves as a short circuit under steady-state DC because $di/dt=0$ .

Capacitance: Capacitance is the ability of a device to store electric charge. Its SI unit is the farad $(F)$ . Its current-voltage relation is

$i_C(t)=C\frac{dv(t)}{dt}$

An ideal capacitor behaves as an open circuit under steady-state DC because $dv/dt=0$ .

Thus, resistors dissipate energy, whereas ideal inductors and capacitors store energy.

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Question 2

Explain the concepts of voltage, current, electrical power, and electrical energy. A device operates at $12\,V$ and draws $2\,A$ for $3$ minutes. Calculate the power and energy consumed.

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Answer 2

Voltage: Voltage is the electrical potential difference between two points. It represents work done per unit charge:

$V=\frac{W}{Q}$

Current: Current is the rate of flow of electric charge:

$I=\frac{dQ}{dt}$

Power: Electrical power is the rate at which electrical energy is transferred or consumed:

$P=VI$

For a resistor, it may also be expressed as

$P=I^2R=\frac{V^2}{R}$

Energy: Electrical energy is power consumed over time:

$E=Pt$

For the given device,

$P=12\times 2=24\,W$

The operating time is

$t=3\times 60=180\,s$

Therefore,

$E=24\times 180=4320\,J$

In watt-hours,

$E=24\times\frac{3}{60}=1.2\,Wh$

Hence, the device consumes $24\,W$ of power and $4320\,J$ or $1.2\,Wh$ of energy.

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Question 3

State and explain Ohm's law. Mention its limitations and calculate the current through a $10\,\Omega$ resistor connected across a $50\,V$ supply.

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Answer 3

Ohm's law: At constant temperature and other physical conditions, the current through a conductor is directly proportional to the voltage across it:

$V\propto I$

Introducing the proportionality constant $R$ ,

$V=IR$

where $V$ is voltage, $I$ is current, and $R$ is resistance.

For the given circuit,

$I=\frac{V}{R}=\frac{50}{10}=5\,A$

Therefore, the current is $5\,A$ .

Limitations of Ohm's law:

It is valid only when temperature and physical conditions remain constant.
It is not directly applicable to nonlinear elements such as diodes and transistors.
It does not describe devices whose resistance changes with voltage or current.
It is not valid for electrolytes, electric arcs, and gas-discharge devices under all conditions.

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Question 4

State Kirchhoff's Current Law and Kirchhoff's Voltage Law. Explain their physical basis and application in circuit analysis.

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Answer 4

Kirchhoff's Current Law (KCL): The algebraic sum of currents at any node is zero:

$\sum I=0$

Equivalently, the total current entering a node equals the total current leaving it:

$\sum I_{\text{in}}=\sum I_{\text{out}}$

KCL is based on the conservation of electric charge. It is mainly used in nodal analysis.

Kirchhoff's Voltage Law (KVL): The algebraic sum of all voltages around any closed loop is zero:

$\sum V=0$

Equivalently, the sum of voltage rises equals the sum of voltage drops in a closed loop. KVL is based on the conservation of energy and is mainly used in mesh or loop analysis.

Sign convention:

While crossing a source from negative to positive, take a voltage rise.
While crossing a resistor in the direction of current, take a voltage drop $IR$ .
A consistent sign convention must be followed throughout the equation.

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Question 5

Describe the basic systematic method and the intuitive method of DC circuit analysis. Compare their advantages and limitations.

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Answer 5

Basic systematic method: This method applies established circuit laws and algebraic procedures. The usual steps are:

Identify all nodes, branches, loops, sources, and element values.
Assign reference directions to branch currents and polarities to voltages.
Simplify series and parallel combinations where possible.
Apply Ohm's law, KCL, and KVL.
Use nodal, mesh, or network-theorem methods to form equations.
Solve the equations and verify power balance.

Intuitive method: This method uses circuit inspection and physical reasoning. Examples include recognizing equal current in series elements, equal voltage across parallel elements, symmetry, voltage division, current division, and open- or short-circuit behavior.

Comparison:

The systematic method is reliable for large or unfamiliar networks.
The intuitive method is faster for simple and symmetric circuits.
Systematic analysis gives a clear mathematical procedure but may require many equations.
Intuitive analysis requires experience and may not work efficiently for complex networks.
In practice, intuitive simplification is usually performed first, followed by a systematic method if necessary.

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Question 9

Explain star-delta and delta-star transformations. Derive the conversion formulas and convert a balanced delta having $3\,\Omega$ in each branch into an equivalent star.

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Answer 9

Star-delta transformations are used when a resistor network cannot be simplified directly using series and parallel combinations.

Let a delta network contain $R_{AB}$ , $R_{BC}$ , and $R_{CA}$ . The equivalent star resistances are

$R_A=\frac{R_{AB}R_{CA}}{R_{AB}+R_{BC}+R_{CA}}$

$R_B=\frac{R_{AB}R_{BC}}{R_{AB}+R_{BC}+R_{CA}}$

$R_C=\frac{R_{BC}R_{CA}}{R_{AB}+R_{BC}+R_{CA}}$

Let the star resistances be $R_A$ , $R_B$ , and $R_C$ . The equivalent delta resistances are

$R_{AB}=R_A+R_B+\frac{R_AR_B}{R_C}$

$R_{BC}=R_B+R_C+\frac{R_BR_C}{R_A}$

$R_{CA}=R_C+R_A+\frac{R_CR_A}{R_B}$

These formulas are obtained by equating the resistance measured between every pair of external terminals in the two networks.

For a balanced delta,

$R_{AB}=R_{BC}=R_{CA}=3\,\Omega$

Therefore, each equivalent star resistance is

$R_A=R_B=R_C=\frac{3\times 3}{3+3+3}=1\,\Omega$

Hence, a balanced $3\,\Omega$ delta is equivalent to a balanced star with $1\,\Omega$ in each arm.

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Question 10

Distinguish between independent and dependent sources. Describe the four types of dependent sources used in electrical circuits.

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Answer 10

Independent source: An independent voltage or current source supplies a specified voltage or current that does not depend on any other circuit variable. Examples are an ideal $10\,V$ voltage source and an ideal $2\,A$ current source.

Dependent source: A dependent or controlled source has a value determined by another voltage or current elsewhere in the circuit. It is represented by a diamond-shaped symbol.

The four types are:

Voltage-Controlled Voltage Source (VCVS):
$v_s=\mu v_x$
where $\mu$ is a dimensionless voltage gain.
Current-Controlled Voltage Source (CCVS):
$v_s=r_m i_x$
where $r_m$ has the unit of ohms.
Voltage-Controlled Current Source (VCCS):
$i_s=g_m v_x$
where $g_m$ has the unit of siemens.
Current-Controlled Current Source (CCCS):
$i_s=\beta i_x$
where $\beta$ is a dimensionless current gain.

Dependent sources model active devices such as transistors and operational amplifiers. They must normally remain active when applying superposition or finding equivalent resistance.

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Question 11

Explain mesh analysis and solve a two-mesh circuit whose mesh equations are $6I_1-2I_2=10$ and $-2I_1+8I_2=5$ . Also find the current through the common branch.

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Answer 11

Mesh analysis is a systematic method based on KVL. It is applicable to planar circuits.

Procedure:

Identify all independent meshes.
Assign a mesh current, usually clockwise, to each mesh.
Apply KVL to every mesh.
For a shared resistor $R$ , write its voltage drop in mesh 1 as $R(I_1-I_2)$ .
Solve the simultaneous equations.

The given equations are

$6I_1-2I_2=10$

$-2I_1+8I_2=5$

Multiplying the first equation by $4$ gives

$24I_1-8I_2=40$

Adding the second equation gives

$22I_1=45$

Therefore,

$I_1=\frac{45}{22}\,A\approx 2.045\,A$

Substituting into the first equation,

$I_2=\frac{25}{22}\,A\approx 1.136\,A$

If the mesh currents pass through the common branch in opposite directions, its current is

$I_{common}=I_1-I_2=\frac{20}{22}=\frac{10}{11}\,A$

Therefore, the common-branch current is approximately $0.909\,A$ in the direction of $I_1$ .

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Question 12

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Answer 12

Nodal analysis determines unknown node voltages by applying KCL.

Procedure:

Select a reference or ground node.
Assign voltage variables to all other essential nodes.
Express branch currents using Ohm's law.
Apply KCL at each unknown node.
Solve the simultaneous equations.

At node $A$ ,

$\frac{V_A-12}{3}+\frac{V_A}{6}+\frac{V_A-V_B}{2}=0$

Multiplying by $6$ ,

$2(V_A-12)+V_A+3(V_A-V_B)=0$

Therefore,

$2V_A-V_B=8$

At node $B$ ,

$\frac{V_B}{4}+\frac{V_B-V_A}{2}=0$

Multiplying by $4$ ,

$3V_B-2V_A=0$

Thus,

$V_B=\frac{2V_A}{3}$

Substitution into the first equation gives

$2V_A-\frac{2V_A}{3}=8$

$V_A=6\,V$

Therefore,

$V_B=\frac{2\times 6}{3}=4\,V$

Hence, $V_A=6\,V$ and $V_B=4\,V$ .

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Question 13

Compare mesh analysis and nodal analysis. State the situations in which each method is preferable.

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Answer 13

Mesh analysis:

Uses KVL to determine mesh currents.
The number of equations equals the number of independent meshes.
It is directly applicable only to planar circuits.
It is usually convenient when the circuit has fewer meshes than nodes.
Voltage sources are handled easily.
A current source common to two meshes requires a supermesh.

Nodal analysis:

Uses KCL to determine node voltages.
The number of equations is generally one less than the number of nodes.
It can be applied to both planar and non-planar circuits.
It is usually convenient when the circuit has fewer essential nodes than meshes.
Current sources are handled easily.
A voltage source between two non-reference nodes requires a supernode.

Selection guideline: Choose mesh analysis when the required quantities are loop currents and the network contains many voltage sources. Choose nodal analysis when node voltages are required or when the network contains many current sources. The method that produces fewer simultaneous equations is generally preferable.

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Question 14

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Answer 14

Thevenin's theorem: Any linear bilateral two-terminal network containing sources and resistances can be replaced by an equivalent voltage source $V_{Th}$ in series with a resistance $R_{Th}$ .

Procedure:

Remove the load connected across the required terminals.
Find the open-circuit terminal voltage; this is $V_{Th}$ .
Deactivate all independent sources: short independent voltage sources and open independent current sources.
Find the resistance seen from the terminals; this is $R_{Th}$ .
Reconnect the load to the Thevenin equivalent.

The open-circuit voltage is obtained by voltage division:

$V_{Th}=12\frac{8}{4+8}=8\,V$

After short-circuiting the independent voltage source, the two resistors appear in parallel:

$R_{Th}=4\parallel 8=\frac{4\times 8}{4+8}=\frac{8}{3}\,\Omega$

For $R_L=4\,\Omega$ , the load current is

$I_L=\frac{V_{Th}}{R_{Th}+R_L}$

$I_L=\frac{8}{\frac{8}{3}+4}=1.2\,A$

The load voltage is

$V_L=I_LR_L=1.2\times 4=4.8\,V$

Thus, the Thevenin equivalent is an $8\,V$ source in series with $8/3\,\Omega$ , and the load current is $1.2\,A$ .

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Question 15

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Answer 15

Norton's theorem: Any linear bilateral two-terminal network can be replaced by an equivalent current source $I_N$ in parallel with an equivalent resistance $R_N$ .

The Norton current is the short-circuit current at the terminals:

$I_N=I_{SC}$

The Norton resistance is the resistance seen from the terminals after deactivating independent sources. The relationships with the Thevenin equivalent are

$R_N=R_{Th}$

$I_N=\frac{V_{Th}}{R_{Th}}$

$V_{Th}=I_NR_N$

For the given network,

$I_N=\frac{8}{8/3}=3\,A$

$R_N=\frac{8}{3}\,\Omega$

Thus, the Norton equivalent is a $3\,A$ source in parallel with $8/3\,\Omega$ .

With $R_L=4\,\Omega$ , current division gives

$I_L=I_N\frac{R_N}{R_N+R_L}$

$I_L=3\frac{8/3}{8/3+4}=1.2\,A$

Hence, the current through the $4\,\Omega$ load is $1.2\,A$ .

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Question 16

State and derive the maximum power transfer theorem for a DC resistive circuit. Also determine the maximum power delivered by a Thevenin source having $V_{Th}=12\,V$ and $R_{Th}=3\,\Omega$ .

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Answer 16

Maximum power transfer theorem: A load receives maximum power from a linear DC network when the load resistance equals the Thevenin resistance seen from the load terminals:

$R_L=R_{Th}$

For a Thevenin equivalent, the load current is

$I=\frac{V_{Th}}{R_{Th}+R_L}$

The load power is

$P_L=I^2R_L=\frac{V_{Th}^2R_L}{(R_{Th}+R_L)^2}$

Differentiate with respect to $R_L$ and equate the result to zero:

$\frac{dP_L}{dR_L}=0$

This gives

$R_L=R_{Th}$

The maximum power is therefore

$P_{max}=\frac{V_{Th}^2}{4R_{Th}}$

For $V_{Th}=12\,V$ and $R_{Th}=3\,\Omega$ ,

$R_L=3\,\Omega$

$P_{max}=\frac{12^2}{4\times 3}=12\,W$

At maximum power transfer, the efficiency is only $50\%$ because the power dissipated in $R_{Th}$ equals the power delivered to $R_L$ .

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Question 17

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Answer 17

Superposition theorem: In a linear circuit containing multiple independent sources, the current or voltage in any element equals the algebraic sum of the responses produced by each independent source acting alone.

When considering one source:

Replace other independent voltage sources by short circuits.
Replace other independent current sources by open circuits.
Keep dependent sources active.

Effect of the $12\,V$ source alone: The conductance connected to the node is

$G=\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1\,S$

The contribution to node voltage is

$V_1=\frac{12/6}{1}=2\,V$

Effect of the $6\,V$ source alone:

$V_2=\frac{6/3}{1}=2\,V$

By superposition,

$V=V_1+V_2=2+2=4\,V$

Therefore, the node voltage is $4\,V$ .

Superposition applies to voltages and currents, but power cannot be directly superimposed because power is a nonlinear function such as $I^2R$ or $V^2/R$ .

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Question 18

Explain source transformation. Convert a $12\,V$ voltage source in series with $4\,\Omega$ into an equivalent current source, and show that both forms produce the same current in an $8\,\Omega$ load.

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Answer 18

Source transformation converts a practical voltage source into an equivalent practical current source, or vice versa, without changing the terminal behavior.

A voltage source $V_s$ in series with $R_s$ is equivalent to a current source

$I_s=\frac{V_s}{R_s}$

in parallel with the same resistance $R_s$ .

For the given source,

$I_s=\frac{12}{4}=3\,A$

Thus, the equivalent is a $3\,A$ current source in parallel with $4\,\Omega$ .

Using the voltage-source form with $R_L=8\,\Omega$ ,

$I_L=\frac{12}{4+8}=1\,A$

Using the current-source form and current division,

$I_L=3\frac{4}{4+8}=1\,A$

Both source forms produce the same load current of $1\,A$ and the same load voltage of

$V_L=I_LR_L=8\,V$

Therefore, the two sources are terminally equivalent.

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Question 20

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Answer 20

Dependent sources are analyzed using KCL, KVL, nodal analysis, or mesh analysis in the same way as independent sources. However, the controlling relationship of the dependent source must be included as an additional circuit equation.

Applying KCL at the node, the current leaving through the resistor equals the total current injected by the sources:

$\frac{v}{5}=2+0.1v$

Since $1/5=0.2$ ,

$0.2v=2+0.1v$

$0.1v=2$

Therefore,

$v=20\,V$

The resistor current is

$I_R=\frac{v}{5}=\frac{20}{5}=4\,A$

The dependent-source current is

$I_d=0.1v=0.1\times 20=2\,A$

The total injected current is therefore $2+2=4\,A$ , which satisfies KCL.

The power absorbed by the resistor is

$P_R=\frac{v^2}{R}=\frac{20^2}{5}=80\,W$

Hence, the node voltage is $20\,V$ , and the resistor absorbs $80\,W$ .

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Unit2

Unit 1: Fundamentals of D.C. circuits - Subjective Questions

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