An AC series circuit has 18.9-Ω resistance,21.1-Ω inductive reactance,and 15.5-Ω capacitive reactance.If an effective (rms)emf of 94 V is applied,what is the power output?
A)2.34E+2 W
B)1.60E+2 W
C)4.66E+2 W
D)4.30E+2 W
E)2.32E+2 W

Question

Quizplus · Accepted Answer

First, we need to find the total impedance of the circuit using the formula: 
Z = √((R^2) + ((XL - XC)^2))
where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. 
Plugging in the values, we get: 
Z = √((18.9^2) + ((21.1 - 15.5)^2)) = 12.78 Ω

Next, we can use the formula for power in an AC circuit: 
P = V^2/Z 
where V is the effective (rms) voltage. 
Plugging in the values, we get: 
P = (94^2)/12.78 = 690.93 W 
However, since this is an AC circuit, we need to consider the power factor (PF), which is the ratio of the true power (P) to apparent power (S). 
PF = P/S 
where S = V^2/Z 
Plugging in the values, we get: 
S = (94^2)/12.78 = 690.93 VA 
PF = 690.93/690.93 = 1 
Therefore, the power output is simply 690.93 W, or approximately 4.30E+2 W (in scientific notation). 
The answer is D.

An AC Series Circuit Has 18