Question

A 200-Ω resistor is connected in series with a 10-µF capacitor and a 60-Hz, 120-V (rms) line voltage. If electrical energy costs 5.0¢ per kWh, how much does it cost to leave this circuit connected for 24 hours?

Answer 1

Answer:

Cost to leave this circuit connected for 24 hours is $ 3.12.

Explanation:

We know that,

$\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}$

f = frequency (60 Hz)

c= capacitor (10 µF = $10^-6$)  

$\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }$

Substitute the given values

$\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}$

$\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}$

$\mathrm{x}_{\mathrm{c}}=265.39 \Omega$

Given that, R = 200 Ω

$X^{2}=R^{2}+X c^{2}$

$X^{2}=200^{2}+265.39^{2}$

$X^{2}=40000+70431.85$

$X^{2}=110431.825$

$x=\sqrt{110431.825}$

X = 332.31 Ω

$\text { Current }(I)=\frac{V}{R}$

$\text { Current }(I)=\frac{120}{332.31}$

Current (I) = 0.361 amps

“Real power” is only consumed in the resistor,  

$\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200$

$\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200$

$\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}$

In one hour 26 watt hours are used.

Energy used in 54 hours = 26 × 24 = 624 watt hours

E = 0.624 kilowatt hours

Cost = (5)(0.624) = 3.12