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Chapter 32: Q. 38 (page 925)

A series RLC circuit attached to a 120V/60Hzpower line draws a 2.4Arms current with a power factor of 0.87. What is the value of the resistor?

Short Answer

Expert verified

The value of resistor is43.5惟.

Step by step solution

01

Given Information

We are given that rms current I=2.4A,Voltage is V=120Vand power factor is Pf=0.87

and we have to find out the value of resistor.

02

Formula Used

Now, using the formula,

We get P=(V)(I)(Pf)

where V is voltage, I is current and Pfis the power factor.

Putting the value of variables in equation,

We get,

P=(120)(2.4)(0.87)=250.56W

03

Simplify

Now, using the formula

P=I2R,

where P is power, I is current and R is resistor,

Putting all the variable in equation we get,

250.56=(2.4)2R

On solving,

we get

R=43.5惟

The value of resistor is43.5惟.

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a. What is the peak current supplied by the emf in FIGUREP32.45?

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the peak volt-age at 250kHZis 2.2V.

a. What is the capacitance?

b. If the peak voltage is held constant, what is the peak current at 500kHZ?

Lightbulbs labeled 40W,60Wand 100W are connected to a120V/60Hzpower line as shown in the figure .What is the rate at which energy is dissipated in each bulb?

a. Show that the average power loss in a series RLC circuit is

Pavg=蝇2蔚rms2R蝇2R2+L2蝇2-蝇o22

b. Prove that the energy dissipation is a maximum at蝇=蝇o.
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