The impedance of the circuit shown is:
A) 21 $\Omega$
B) 50 $\Omega$
C) 63 $\Omega$
D) 65 $\Omega$
E) 98 $\Omega$

Question

The impedance of the circuit shown is: 
A) 21 $\Omega$ 
B) 50 $\Omega$ 
C) 63 $\Omega$ 
D) 65 $\Omega$ 
E) 98 $\Omega$

Quizplus · Accepted Answer

To calculate the impedance of the circuit, we need to sum up the impedance of each element.

For the resistor: 
Zr = R = 40 Ω

For the inductor: 
ZL = jωL = j(100 × 10^3 × 0.5 × 10^(-6)) = j0.05 Ω

For the capacitor: 
ZC = 1/(jωC) = 1/(j100 × 10^3 × 1 × 10^(-6)) = -j10 Ω

The negative sign in the impedance of the capacitor indicates that the capacitor acts as a reactive element that opposes the flow of current.

Now, we can sum up the impedances:

Z = Zr + ZL + ZC = (40-j10) + j0.05 - j10 = 30-j9.95

The magnitude of the impedance is: 
|Z| = √(30^2 + (-9.95)^2) = 31.13 Ω

Rounding this to the nearest integer, we get 31 Ω, which in scientific notation is 31 × 10^(0) Ω = 31 Ω.

Therefore, the closest option is D) 65 Ω.

The Impedance of the Circuit Shown Is: A) 21 Ω\OmegaΩ