A $ 2700\Omega $ resistor and a $ 1.1\mu F $ capacitor are connected in series across a generator (60 Hz, 120 V). Determine the power consumed in the circuit.
A) 3 W
B) 9 W
C) 10 W
D) 7 W

Hint : Since the generator delivers an AC supply to the circuit, the power delivered will be due to the resistance and the capacitive reactance. We will calculate the capacitive reactance for given circuit parameters and then calculate the power delivered to the circuit.

Formula used: In this question, we will use the following formula
 $ {X_C} = \dfrac{1}{{2\pi fC}} $ where $ {X_C} $ is the capacitive reactance, $ f $ is the frequency of the power supply, and $ C $ is the value of the capacitance

Complete step by step answer
We’ve been given that a $ 2700\Omega $ resistor and a $ 1.1\mu F $ capacitor are connected in series across a generator which provides an AC supply. The capacitive reactance of the circuit can be calculated by using the formula
 $ {X_C} = \dfrac{1}{{2\pi fC}} $
Substituting $ f = 60\,Hz $ and $ C = 1.1 \times {10^{ - 6}}\,\mu F $ , we get
 $ {X_C} = \dfrac{1}{{2\pi (60)(1.1 \times {{10}^{ - 6}})}} $
 $ \Rightarrow {X_C} = 2400\,\Omega $
The average power delivered in the circuit will now be due to the resistor as well as the capacitive reactance and can be calculated as
 $ P = \dfrac{{{V^2}R}}{{{R^2} + {X_C}^2}} $
Substituting $ V = 120\, $ , we can calculate the average power delivered as
 $ P = \dfrac{{{{(120)}^2}2700}}{{{{(2700)}^2} + {{(2400)}^2}}} $
 $ \Rightarrow P = 3\,W $
Hence $ 3\,W $ power will be delivered to the circuit by the generator which corresponds to option (A).

Note
The power delivered to the circuit that we just calculated will be delivered to the circuit for one cycle of the input power. We must carefully notice that the generator provides an AC supply to the circuit since it has a frequency that has been provided to us. If the supply was DC, the power delivered in the circuit would just be across the resistor depending on the current in the circuit and would be exponentially decaying in time once the capacitor starts charging.