[amp_mcq option1=”4 kVAR” option2=”6 kVAR” option3=”circuit is 10 kVA and active power is 8 kW. The reactive power in the circuit is A. 4 kVAR B. 6 kVAR C. 8 kVAR” option4=”16 kVAR” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{B. }6 \\text{ kVAR}}$.<\/p>\n
The apparent power is the product of the voltage and current in the circuit, and is given by the formula $S = \\sqrt{P^2 + Q^2}$, where $P$ is the active power and $Q$ is the reactive power. The active power is the power that is actually used by the circuit, and is measured in watts (W). The reactive power is the power that is stored and released by the circuit’s inductance and capacitance, and is measured in vars (VAr).<\/p>\n
In this case, the apparent power is 10 kVA and the active power is 8 kW. Substituting these values into the formula, we get $10^2 = 8^2 + Q^2$. Solving for $Q$, we get $Q = \\sqrt{10^2 – 8^2} = \\sqrt{36} = 6$ kVAR.<\/p>\n
Therefore, the reactive power in the circuit is 6 kVAR.<\/p>\n
Option A is incorrect because it is the active power, not the reactive power. Option C is incorrect because it is the apparent power, not the reactive power. Option D is incorrect because it is twice the reactive power, which is not possible.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”4 kVAR” option2=”6 kVAR” option3=”circuit is 10 kVA and active power is 8 kW. The reactive power in the circuit is A. 4 kVAR B. 6 kVAR C. 8 kVAR” option4=”16 kVAR” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[739],"tags":[],"class_list":["post-58495","post","type-post","status-publish","format-standard","hentry","category-a-c-fundamentals-circuits-and-circuit-theory","no-featured-image-padding"],"yoast_head":"\n