Mastering Circuits – Basic Concepts of Electrical Circuits (14 Solved Problems)

Mastering Circuits!

Mastering the basic concepts of electrical circuits is the foundation for every student and aspiring engineer. In this comprehensive guide, we break down the essential circuit theory fundamentals, including SI Units, electric charge, resistance, current, and voltage. You will also explore the relationship between power and energy, and learn how to identify various circuit elements. Whether you are preparing for an exam or refreshing your knowledge, join Mastering Circuits as we solve practical problems to help you bridge the gap between theory and practice.

Introduction

An electric circuit is an interconnection of electrical elements.

Figure 1 – A Simple Electric Circuit

A simple electric circuit is shown in Fig. 1. It consists of three basic elements: a battery, a lamp, and connecting wires.

In electrical engineering, we are often interested in communicating or transferring energy from one point to another point. To do this requires an interconnection of electrical devices. Such interconnection is referred to as an electric circuit, and each component is known as an element.

System of Units

To analyze circuits accurately, we use the International System of Units (SI). Understanding these base units and their prefixes is essential for solving engineering problems.

Basic SI Units

Physical Quantity	Si Unit	Symbol
Length	meter	m
Mass	kilogram	kg
Time	second	s
Temperature	kelvin	K
Electrical Current	ampere	A
Light Intensity	candela	cd
Amount of Substance	mole	mol

The SI Prefixes

Multiplier	Prefix	Symbol
10-1	deci	d
10-2	centi	c
10-3	milli	m
10-6	micro	µ
10-9	nano	n
10-12	pico	p
10-15	femto	f
10-18	atto	a
101	deka	da
102	hecto	h
103	kilo	k
106	mega	M
109	giga	G
1012	tera	T
1015	peta	P
1018	exa	E

Charge, Current & Resistance

Charge: Charge is an electrical property of the atomic particles of which matter consists, measured in coulombs (C).

Matter is made of atoms and each atom consists of electrons, protons and neutrons.

e = -1.602 \times 10^{-19} C p = +1.602 \times 10^{-19} C

Current: Electric current is the time rate of change of charge, measured in amperes (A).

i = \frac{dq}{dt}; \quad 1 A = 1 C/s

Figure 2 – Electric Current due to flow of electronic charge in a conductor

• Current flows from positive to negative.
• Electronic charge flows from negative to positve.

Q = \int_{t_{0}}^{t} i \, dt

Current: 2 types

1. DC (Direct Current)
2. AC (Alternating Current)

A direct current (DC) flows only in one direction and can be constant or time varying.

Figure 3 – Direct Current (DC)

An alternating current (AC) is a current that changes direction with respect to time.

Figure 4 – Alternating Current (AC)

Figure 5 – Conventional current flow : (a) positive current flow, (b) negative current flow

As current is defined as the movement of charge, so current will have an associated direction of flow.

The direction of current flow is conventionally taken as the direction of positive charge movement.

Based on this convention, a current of 5 A is represented positively or negatively as shown in Fig. 5.

A negative current of -5 A flowing in one direction is the same as a current of 5 A flowing in the opposite direction.

Resistance: Resistance is a measure of the opposition to current flow in an electrical circuit, measured in ohms ( \Omega ).

All materials resist current flow to some degree. They fall into one of two broad categories:

• Conductors: Materials that offer very little resistance where electrons can move easily. Ex: Silver, Copper, Gold, Aluminium.
• Insulators: Materials that present high resistance and restrict the flow of electrons. Ex: Rubber, Paper, Glass, Wood, Plastic.

Figure 6 – Resistivity of a conductor

Figure 7 – Circuit symbol for resistance

Resistance is the ability of an element to resist electric current, measured in ohms ( \Omega ).

Solved Problems on Charge, Current & Resistance

Problem 1

Pb-1: How much charge is represented by 4,600 electrons?

Solution:

\begin{aligned} 1e &= -1.602 \times 10^{-19} \, C \\ \therefore 4600e &= \{(-1.602 \times 10^{-19} \, C) \times 4600\} \\ &= -7.369 \times 10^{-16} \, C \end{aligned}

Problem 2

Pb-2: Calculate the amount of charge by 10 billion protons.

Solution:

\begin{aligned} 1 \, crore &= 1,00,00,000 = 10^7 \\ 1 \, billion &= 100 \, crores = 10^9 \\ \therefore 10 \, billion &= 10^{10} \\ \small \therefore 10 \, billion \, protons & \small = 10^{10} \times 1.602 \times 10^{-19} \, C \\ &= 1.602 \times 10^{-9} \, C \end{aligned}

Problem 3

Pb-3: The total charge entering a terminal is given by q = 5t \sin 4\pi t \, mC . Calculate the current at t = 0.5s .

Solution:

i = \frac{dq}{dt} = \frac{d}{dt} (5t \sin 4\pi t) \bigg|_{t=0.5} = 31.42 \, mA

Problem 4

Pb-4: The total charge entering a terminal is given by q = (20 - 15t - 10e^{-3t}) \, mC . Find the current at t = 0.5s .

Solution:

\footnotesize i = \frac{dq}{dt} = \frac{d}{dt} (20 - 15t - 10e^{-3t}) \bigg|_{t=1.0} = -13.506 \, mA

Problem 5

Pb-5: Determine the total charge entering a terminal between t = 1s and t = 2s if the current passing the terminal is i = (3t^{2} - t)A .

Solution:

Q = \int_{t_{1}}^{t_{2}} i \, dt = \int_{1}^{2} (3t^{2} - t) \, dt = 5.5 \, C

Problem 6

Pb-6: The current flowing through an element is

i = \begin{cases} 8 \, A, & 0 < t < 1 \\ 8t^{2} \, A, & t > 1 \end{cases}

Calculate the charge entering the element from t = 0s to t = 2s .

Solution:

\begin{aligned} Q = \int_{t_1}^{t_2} i \, dt &= \int_{0}^{2} i \, dt = \int_{0}^{1} i \, dt + \int_{1}^{2} i \, dt \\ &\small = \int_{0}^{1} 8 \, dt + \int_{1}^{2} 8t^{2} \, dt = 26.67 \, C \end{aligned}

Voltage

Voltage: Voltage (or potential difference) is the energy required to move a unit charge from a reference ( - ) to another point ( + ), measured in Volts ( V ).

v = \frac{dw}{dq}; \qquad 1V = 1J/C = 1N\text{-}m/C

Figure 8 – Polarity of voltage v_{ab}

Fig. 8 shows the voltgae across an element (represented by a rectangular block) connected to points a and b . The plus (+) and minus (-) signs are used to define reference direction or voltage polarity.

v_{ab} = -v_{ba}

Figure 9 – Two equivalent representations of the same voltage v_{ab} : (a) Point a is 9 V above point b ; (b) Point b is -9 V above point a

• Electric current is always through an element.
• Electric voltage is always across the element or between two points.

Power & Energy

Power: Power is the time rate of expanding or absorbing energy, measured in watts (W) .

\begin{array}{l|l} p = \frac{dw}{dt} & \text{Where,} \\ \Rightarrow p = \frac{dw}{dq} \cdot \frac{dq}{dt} = vi & v = \frac{dw}{dq}; \, i = \frac{dq}{dt} \\ \therefore p = vi \end{array}

The power p = vi is a time-varying quantity and is called instantaneous power.

If the power has a + sign, power is being delivered or absorbed by the element.

If the power has a - sign, power is being supplied by the element.

Figure 10 – Reference polarities for power using the passive sign convention : (a) absorbing power, (b) supplying power

Passive sign convention is satisfied when the current enters through the positive terminal of an element and p = +vi . If the current enters through the negative terminal, p = -vi .

Figure 11 – Two cases of an element with an absorbing power of 12 W : (a) p = 4\times3 = 12 W , (b) p = 4\times3 = 12 W

Figure 12 – Two cases of an element with a supplying power of 12 W : (a) p = -4\times3 = -12 W , (b) p = -4\times3 = -12 W

+ Power absorbed = - Power Supplied

The algebraic sum of power in a circuit at any instant of time must be zero.

\sum p = 0

Energy: Energy is the capacity to do work, measured in joules (J) .

\begin{gathered} w = pt = vit \\ w = \int_{t_{0}}^{t} p \, dt = \int_{t_{0}}^{t} vi \, dt \end{gathered}

The electric power utlity companies measure energy in watt-hours (Wh) , where –

1 Wh = 3600 J

Solved Problems on Voltage, Power & Energy

Problem 7

Pb-7: An energy source forces a constant current of 2 A for 10 s to flow through a light bulb. If 2.3 kJ is given off in the form of light and heat energy, calculate the voltage drop across the bulb.

Solution:

\footnotesize \begin{array}{l|l} v = \frac{dw}{dq} = \frac{dw}{idt} = \frac{2.3 \times 10^{3}}{2 \times 10} = 115V & \text{Where,} \\ & i = \frac{dq}{dt} \therefore dq = idt \end{array}

Problem 8

Pb-8: To move charge q from point b to point a requires 100 J . Find the voltage drop v_{ab} (the voltage at a positive with respect to b ) if: (a) q = 5 C , (b) q = -10 C .

Solution:

(a) v_{ab} = \frac{dw}{dq} = \frac{100}{5} = 20V

(b) v_{ab} = \frac{dw}{dq} = \frac{100}{-10} = -10V

Problem 9

Pb-9: Find the power delivered to an element at t = 3 ms if the current entering its positive terminal is

i = 5 \cos 60\pi t \, A

and the voltage is: (a) v = 3i , (b) 3 di/dt

Solution:

(a) \begin{aligned} p &= vi = 3i \times i = 3i^{2} = 3 \times 25 \cos^{2} 60\pi t \\ &\small = 75 \cos^{2}(60\pi \times 3 \times 10^{-3}) = 53.47 \, W \end{aligned}

(b) \begin{aligned} p &= vi = 3\frac{di}{dt} \times i \\ &= 3 \times \frac{d}{dt} (5 \cos 60\pi t) \times 5 \cos 60\pi t \\ &= 3 \times 5 \times (-\sin 60\pi t) \times 60\pi \times 5 \cos 60\pi t \\ &= -4500\pi \sin 60\pi t \cos 60\pi t \\ &\scriptsize = -4500\pi \times \sin(60\pi \times 3 \times 10^{-3}) \times \cos(60\pi \times 3 \times 10^{-3}) \\ &= -6395.85 \, W \end{aligned}

Problem 10

Pb-10: Find the power delivered to an element at t = 5 ms if the current entering its positive terminal is

i = 5 \cos 60\pi t \, A

and the voltage is: \small \begin{aligned} \text{(a)} \quad v &= 6i \\ \text{(b)} \quad v &= \left( 6 + 10 \int_{0}^{t} i \, dt \right) \end{aligned}

Solution:

(a) \begin{aligned} p &= vi = 6i^2 = 6 \times 25 \times \cos^2 60\pi t \\ &\small = 150 \times \cos^2(60\pi \times 5 \times 10^{-3}) \\ &= 51.82 \, W \end{aligned}

(b) \begin{aligned} v &= \left( 6 + 10 \int_{0}^{t} i \, dt \right) \\ &= \left( 6 + 10 \int_{0}^{5 \times 10^{-3}} (5 \cos 60\pi t) \, dt \right) \\ &= 6.21 \, V \\ p &= vi = 6.21 \times 5 \cos 60\pi t \\ &= 31.05 \times \cos(60\pi \times 5 \times 10^{-3}) \\ &\small = 18.251 \, W \end{aligned}

Problem 11

Pb-11: How much energy does a 100 W electric bulb consume in two hours?

Solution:

\begin{aligned} w = pt &= 100 \times 2 = 200 \, Wh = 200 \times 3600 \, J \\ \small &= 7,20,000 \, J = 720 \, kJ \end{aligned}

Problem 12

Pb-12: A home electric heater draws 12 A when connected to a 115 V outlet. How much energy is consumed by the heater over a period of 24 hours?

Solution:

\begin{aligned} w = pt = vit &= 115 \times 12 \times 24 = 33120 \, Wh \\ \small &= 33.12 \, kWh \end{aligned}

Circuit Elements

Circuit: Circuit is a network having closed path to flow electric current or magnetic flux. It is an interconnection of circuit elements.

Circuit Elements: 2 types

• Active Circuit Elements

• Passive Circuit Elements

The most important active elements are voltage or current sources that generally deliver power to the circuit connected to them.

Sources: 2 types

1. Independent Sources
2. Dependent Sources

Independent Sources: An ideal independent source is an active element that provides a specified voltage or current that is completely independent of other circuit element.

Figure 13 – Symbols for independent voltage sources : (a) used for constant or time-varying voltage, (b) used for constant voltage (dc)

Figure 14 – Symbol for independent current source

Dependent Sources: An ideal dependent (or controlled) source is an active element in which the source quantity is controlled by another voltage or current.

Dependent sources are usually designated by diamond-shaped symbols.

Figure 15 – Symbols for : (a) dependent voltage source, (b) dependent current source

There are four possible types of dependent sources, namely:

1. A voltage-controlled voltage source (VCVS)
2. A current-controlled voltage source (CCVS)
3. A voltage-controlled current source (VCCS)
4. A current-controlled current source (CCCS)

Figure 16 – The source on the right hand side is a current controlled voltage source

Solved Problems on Circuit Elements

Problem 13

Pb-13: Calculate the power absorbed or supplied by each element in the Fig. 17.

Figure 17 – Circuit diagram for Pb-13

Solution:

\small \begin{aligned} p_{1} &= 20 \times (-5) = -100 \, W \rightarrow \text{Supplied power} \\ p_{2} &= 12 \times (5) = 60 \, W \rightarrow \text{Absorbed power} \\ p_{3} &= 8 \times (6) = 48 \, W \rightarrow \text{Absorbed power} \\ p_{4} &= 8 \times (-0.2 \, I) = 8 \times (-0.2 \times 5) \\ &= -8 \, W \rightarrow \text{Supplied power} \end{aligned}

Problem 14

Pb-14: Calculate the power absorbed or supplied by each element in the Fig. 18.

Figure 18 – Circuit diagram for Pb-14

Solution:

\small \begin{aligned} p_{1} &= 5 \times (-45) = -225 \, W \rightarrow \text{Supplied power} \\ p_{2} &= 2 \times (45) = 90 \, W \rightarrow \text{Absorbed power} \\ p_{3} &= 0.12 \, I \times (20) = 0.12 \times 25 \times 20 \\ &= 60 \, W \rightarrow \text{Absorbed power} \\ p_{4} &= 3 \times (25) = 75 \, W \rightarrow \text{Absorbed power} \end{aligned}

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Congratulations! You’ve successfully navigated the foundational building blocks of Electrical Circuits. This is the first major step toward total mastery of circuit analysis. If any of these concepts, from SI Units to Circuit Elements, still feel a bit confusing, don’t hesitate to reach out. Drop a comment below with your questions, and let’s clear up the confusion together!

Read Also: Basic Laws of Electrical Circuits

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