[amp_mcq option1=”77.7 ohm” option2=”89.3 ohm” option3=”88.8 ohm” option4=”224 ohm E. None of the above” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B) }89.3\\Omega}$.<\/p>\n
The impedance of a parallel circuit is given by the formula:<\/p>\n
$$Z = \\frac{R}{1+\\frac{X_L}{R}}$$<\/p>\n
where $R$ is the resistance, $X_L$ is the inductive reactance, and $Z$ is the impedance.<\/p>\n
In this case, $R=100\\Omega$ and $X_L=200\\Omega$. Substituting these values into the formula, we get:<\/p>\n
$$Z = \\frac{100\\Omega}{1+\\frac{200\\Omega}{100\\Omega}} = \\frac{100\\Omega}{3} = 89.3\\Omega$$<\/p>\n
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.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”77.7 ohm” option2=”89.3 ohm” option3=”88.8 ohm” option4=”224 ohm E. None of the above” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[682],"tags":[],"class_list":["post-19075","post","type-post","status-publish","format-standard","hentry","category-electronic-principles","no-featured-image-padding"],"yoast_head":"\n