An AC generator is shown connected to a 25.0 ohm resistor in Figure 3a. The inte

Question

An AC generator is shown connected to a 25.0 ohm resistor in Figure 3a. The internal resistance of the generator is 5.00 ohms. It is found that when a 5.40 uF is placed in series with the 25.0 ohm resistor (as in Figure 3b) the current in the resistor is maximized. (Assume rms values for current and voltage.) a.) Explain why the addition of the capacitor in figure 3b maximizes the current, and find this maximum current. (hint: consider the generator's coil.) b) What was the current in the resistor in figure 3a? (hint: first find the impedance of the circuit. Show calculations.) c) In the circuit in figure 3a does the current lead or lag the voltage? What is the phase angle? Show calculations. d) In the circuit in figure 3b does the current lead or lag the voltage? What is the phase angle? (hint: you may briefly explain your answers without calcutions)

http://puu.sh/8lGuY.png if you can't see

Explanation / Answer

a) Due to generator coil's there is an inductor in circuit
curent in an RLS circuit is given by: E/(sqrt(R^2+(XL+Xc)^2))
where Xl is impedence due to inductor and Xc due to capacitance
Current is maximum when XL+Xc=0

This is the case which is hapening here Capacitor added here cancels the impedence due to Inductor

PS- this condition is known as Resonance

b)
w=2*pie*f
f=60Hz
w=120*pie
Xc=-j/wC=-491.2j

=> XL=-Xc=491.2j

net impedence: Z=491.2j+25=491.835/_87.08
Current:120/Z=0.2033/_-87.08
Value of current=0.2033A

c) lags by 87.08

d) Since it is a purely resistive Ckt Current is in phase with the voltage
phase angle=0

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