An alternating current supply is connected in series with a resistor of resistance 12\ \Om… — JAMB UTME

An alternating current supply is connected in series with a resistor of resistance $12\ \Omega$ and an inductor of inductive reactance $5\ \Omega$ . What is the total impedance of the circuit?

$13\ \Omega$ Answer
B
$17\ \Omega$
C
$7\ \Omega$
D
$169\ \Omega$

Answer

13\ \Omega

In a series R-L alternating current circuit, the voltage across the resistor is in phase with the current, while the voltage across the inductor leads the current by

90^\circ

. Consequently, resistance and inductive reactance combine vectorially. The total impedance is

Z = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\ \Omega

Step-by-Step Solution

Identify the given values

Resistance

R = 12\ \Omega

, inductive reactance

X_L = 5\ \Omega

These are the given parameters of the series R-L circuit.

Apply the total impedance formula for a series R-L AC circuit

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

In an AC circuit, resistance and reactance are perpendicular vectors (

90^\circ

out of phase).

Substitute the given values and calculate the result

Z = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\ \Omega

Evaluating the square root yields the total opposition to alternating current.

Key Concept

Impedance of Series AC Circuits

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