An AC series circuit consists of a resistor of resistance R = 6\ \Omega connected in serie… — JAMB UTME

An AC series circuit consists of a resistor of resistance $R = 6\ \Omega$ connected in series with an inductor of inductive reactance $X_L = 8\ \Omega$ . What is the total impedance of the circuit?

$10\ \Omega$ Answer
B
$14\ \Omega$
C
$2\ \Omega$
D
$100\ \Omega$

Answer

The total impedance of the circuit is

10\ \Omega

The net impedance

Z

in an AC series circuit containing a resistor and an inductor is given by

Z = \sqrt{R^2 + X_L^2}

. Substituting

R = 6\ \Omega

and

X_L = 8\ \Omega

yields

Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\ \Omega

Step-by-Step Solution

Identify the formula for impedance in an RL series circuit

Z = \sqrt{R^2 + X_L^2}

In an AC circuit, resistance and inductive reactance add vectorially at a

90^\circ

phase angle.

Substitute the given values

R = 6\ \Omega

and

X_L = 8\ \Omega

into the formula

Z = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}

Squaring each component gives the terms required under the radical.

Calculate the square root

Z = 10\ \Omega

The square root of 100 gives the net opposition to current flow (impedance).

Key Concept

Impedance of an RL Series AC Circuit

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