Answer:
Cost to leave this circuit connected for 24 hours is $ 3.12.
Explanation:
We know that,
[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}[/tex]
f = frequency (60 Hz)
c= capacitor (10 µF = [tex]10^-6[/tex])
[tex]\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }[/tex]
Substitute the given values
[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}[/tex]
[tex]\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}[/tex]
[tex]\mathrm{x}_{\mathrm{c}}=265.39 \Omega[/tex]
Given that, R = 200 Ω
[tex]X^{2}=R^{2}+X c^{2}[/tex]
[tex]X^{2}=200^{2}+265.39^{2}[/tex]
[tex]X^{2}=40000+70431.85[/tex]
[tex]X^{2}=110431.825[/tex]
[tex]x=\sqrt{110431.825}[/tex]
X = 332.31 Ω
[tex]\text { Current }(I)=\frac{V}{R}[/tex]
[tex]\text { Current }(I)=\frac{120}{332.31}[/tex]
Current (I) = 0.361 amps
“Real power” is only consumed in the resistor,
[tex]\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200[/tex]
[tex]\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200[/tex]
[tex]\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}[/tex]
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
