What is the impedance of a circuit composed of a 100 ohm resistor connected in parallel with an inductor that has a reactance of 200 W? A. 77.7 ohm B. 89.3 ohm C. 88.8 ohm D. 224 ohm E. None of the above

[amp_mcq option1=”77.7 ohm” option2=”89.3 ohm” option3=”88.8 ohm” option4=”224 ohm E. None of the above” correct=”option3″]

The correct answer is $\boxed{\text{B) }89.3\Omega}$.

The impedance of a parallel circuit is given by the formula:

$$Z = \frac{R}{1+\frac{X_L}{R}}$$

where $R$ is the resistance, $X_L$ is the inductive reactance, and $Z$ is the impedance.

In this case, $R=100\Omega$ and $X_L=200\Omega$. Substituting these values into the formula, we get:

$$Z = \frac{100\Omega}{1+\frac{200\Omega}{100\Omega}} = \frac{100\Omega}{3} = 89.3\Omega$$

Option A is incorrect because it is the impedance of a series circuit, not a parallel circuit. Option C is incorrect because it is the impedance of a purely resistive circuit, not a circuit with both resistance and reactance. Option D is incorrect because it is the impedance of a purely inductive circuit, not a circuit with both resistance and reactance. Option E is incorrect because it is not one of the possible answers.

Exit mobile version