A-Class Visual Design

Bulb Rlc

physics again?

9) what is the peak (maximum) power dissipated by a 150 w light bulb attached to an ac line??

a) 300 w
b) 212 w
c) 150 w
d) 106 w
e) 75 w

10) An RLC series circuit has a 200 ohms resistance, a 3.0-mH inductor, and 100 uF capacitor. What is the resonant freq for this current?

a) 1.8 kHz
b) 1.8MHz
c) 290 Hz
d) 290 kHz
e) 11kHz

I got b for 9 and a for 10, am i right? if not please explain it to me, thanks

I'm a little rusty with this . . .
Method #1
E is *10^

I choose 100 volts as my line voltage
The resistance = v^2/pwr = 100^2/150 =10000/150 = 6.67E+01 ohms. (resistance of the bulb)

Peak voltage is V*sqrt(2) = 100*1.414=141.1 volts.
Calculate peak power = V^2/R = 141.1^2/66.7 =298.4889055 Peak watts

Answer A 300 watts

P.S. this will work for any voltage

Method #2
To get peak voltage you have to multiply the line voltage by Sqrt(2) approx 1.414. Power is voltage squared so you have to multiply the power by (sqrt(2)) squared or 2. RMS power of 150 watts is sqrt(2)*sqrt(2) * 150 = 2*150 =300 watts

Answer b is wrong because you mulitiplied 150 * sqrt(2) =
212watts. Multiply by sqrt(2) again to get 300 watts

Question 2

1/(2pisqrt(LC) =F
1/(2*3.14159265 *sqrt((3E-3)*(100E-6))) = F
1/(6.283185 * sqrt(3E-7) = F
1/(3.44E-03) = F
290.57584 =F
Answer C 290 Hz

Resistance does not play a part in determining the resonant freq . It does play a part in the Q of a circuit.

P.S. Excel formula is
=1/(2*3.14159265*SQRT((3E-3)*
(100E-6)))=290.5758419
All in one cell.

D.bumstead

Viewsonic RLC-018 Projector Lamp

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