A 50-Ω resistor,a 0.1-H inductor,and a 20-µF capacitor are connected in series to a 60-Hz source.The rms current in the circuit is 3 A.Find the rms voltage across the resistor,the inductor,and the capacitor.
A) 150 V,113 V,398 V (all rms)
B) 150 V,113 V,796 V (all rms)
C) 0.06 V,80 V,562 V (all rms)
D) 60 V,80 V,398 V (all rms)

Question

Quizplus · Accepted Answer

We can use the formula for impedance of each component in the circuit:
- Impedance of resistor = R = 50 Ω
- Impedance of inductor = jωL = j(2πfL) = j(2π*60*0.1) Ω = j37.7 Ω
- Impedance of capacitor = 1/(jωC) = 1/(j2πfC) = 1/(j2π*60*20*10^-6) Ω = -j132.7 Ω (note the negative sign, indicating a phase shift of -90 degrees)

The total impedance of the circuit is the sum of the individual impedances:
Z = R + jωL + 1/(jωC)
Z = 50 + j37.7 - j132.7
Z = 50 -j95 Ω

Using Ohm's Law, we can find the rms voltage across each component:
- Voltage across resistor = IR = 3*50 = 150 V (rms)
- Voltage across inductor = I(jωL) = 3*j37.7 = j113 V (rms)
- Voltage across capacitor = I(1/(jωC)) = 3*(-j132.7) = -j398 V (rms)

Note that the voltage across the capacitor is also negative, indicating a phase shift of -90 degrees.

A 50-Ω Resistor,a 0