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Solutions Manual Fundamentals of Electric Circuits 7th Edition Alexander Sadiku

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Master complex engineering circuit analysis with the official Solutions Manual for Fundamentals of Electric Circuits (7th Edition) by Charles K. Alexander and Matthew N.O. Sadiku. This comprehensive academic resource provides step-by-step, fully worked-out solutions for all end-of-chapter problems across the textbook. Perfect for electrical and computer engineering students, it delivers clear mathematical derivations and circuit diagrams covering nodal analysis, mesh analysis, operational amplifiers, AC steady-state analysis, and Laplace transforms.

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Solutions Manual: Fundamentals of Electric Circuits,
7th Edition by Charles K. Alexander & Matthew N.O.
Sadiku – with Complete Solutions




CHAPTER 1: BASIC CONCEPTS (Questions 1-30)




Question 1:
Which of the following is the SI unit of electric charge?

A) Ampere
B) Volt
C) Coulomb
D) Watt

*Rationale: * The coulomb (C) is the SI unit of electric charge, defined as the charge
transported by a constant current of one ampere in one second. One coulomb is
equivalent to approximately 6.24 × 10^18 elementary charges .




Question 2:
What is the SI unit of electric current?

,A) Coulomb
B) Volt
C) Ampere
D) Ohm

*Rationale: * The ampere (A) is the SI unit of electric current, representing the flow of one
coulomb of charge per second. It is one of the seven base SI units .




Question 3:
A current of 5 amperes flowing for 10 seconds transfers how much charge?

A) 2 C
B) 15 C
C) 50 C
D) 0.5 C

*Rationale: * Charge (Q) equals current (I) multiplied by time (t): Q = I × t = 5 A × 10 s =
50 C. This is the fundamental relationship between current, charge, and time.




Question 4:
The voltage across a resistor is 12 V and the current through it is 3 A. What is the
resistance?

A) 36 Ω
B) 15 Ω
C) 4 Ω
D) 0.25 Ω

*Rationale: * Using Ohm's Law, R = V/I = 12 V / 3 A = 4 Ω. This is the foundational
relationship for circuit analysis.




Question 5:
Power absorbed by a resistor can be calculated using which formula?

,A) P = I²R
B) P = V²/R
C) P = VI
D) All of the above

*Rationale: * All three formulas are equivalent for resistors. Using Ohm's Law (V = IR), we
can derive P = VI = I²R = V²/R. These are all valid expressions for power dissipated in a
resistor.




Question 6:
A 100 W light bulb is connected to a 120 V source. What current does it draw?

A) 1.2 A
B) 0.83 A
C) 0.833 A
D) 12 A

*Rationale: * Using P = VI, I = P/V = 100 W / 120 V ≈ 0.833 A. This demonstrates the
practical application of power calculations.




Question 7:
What is the power absorbed by a resistor with resistance 10 Ω when a current of 2 A
flows through it?

A) 20 W
B) 10 W
C) 5 W
D) 40 W

*Rationale: * Using P = I²R = (2 A)² × 10 Ω = 4 × 10 = 40 W. The power is proportional
to the square of the current.

, Question 8:
If the voltage across an element is 8 V and the current entering the positive terminal is 2
A, what is the power?

A) 16 W (absorbed)
B) 16 W (supplied)
C) 4 W
D) 6 W

*Rationale: * When current enters the positive terminal of an element, the element
absorbs power. P = VI = 8 V × 2 A = 16 W absorbed .




Question 9:
If the voltage across an element is 8 V and the current leaves the positive terminal
(negative current entering positive terminal), what is the power?

A) -16 W (supplied)
B) 16 W (absorbed)
C) 4 W
D) -4 W

*Rationale: * When current flows out of the positive terminal, the element supplies power.
The power is negative when using passive sign convention: P = VI = 8 V × (-2 A) = -16 W,
indicating power is being supplied .




Question 10:
Which of the following is NOT a passive element?

A) Resistor
B) Capacitor
C) Inductor
D) Voltage source

*Rationale: * Resistors, capacitors, and inductors are passive elements that absorb or store
energy. Voltage and current sources are active elements that supply energy to the circuit.

Información del documento

Subido en
12 de agosto de 2026
Número de páginas
57
Escrito en
2026/2027
Tipo
Examen
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