When connected in parallel across a 120 V source, two lightbulbs consume 40 W an

Question

When connected in parallel across a 120 V source, two lightbulbs consume 40 W and 60 W, respectively. What powers do the lightbulbs consume if instead they are connected in series across the same source? Assume the resistance of each lightbulb is constant.

P40

P60

2    W

2.If your local power company charges $0.13/kW · h, what would it cost to run a 1300 W heater continuously during a 9-h night?
$ 1

3.A capacitor of 18 F and a resistor of 100 are quickly connected in series to a battery of 6.0 V. What is the charge on the capacitor 0.0010 s after the connection is made?
1 C

4.A 10.0 F capacitor has an initial charge of 100.0 C. If a resistance of 50.0 is connected across it, what is the initial current through the resistor?
1 A

5.13.3 F capacitor is charged to a potential of 55.0 V and then discharged through a 75.0 resistor.

(a) How long after discharge begins does it take for the capacitor to lose 90.0% of the following?

(i) its initial charge
1 s

(ii) its initial energy
2 s

(b) What is the current through the resistor at both times in part (a)?

(i) at tcharge
3 A

(ii) at tenergy
4 A

P40

P60

2    W

2.If your local power company charges $0.13/kW · h, what would it cost to run a 1300 W heater continuously during a 9-h night?
$ 1

3.A capacitor of 18 F and a resistor of 100 are quickly connected in series to a battery of 6.0 V. What is the charge on the capacitor 0.0010 s after the connection is made?
1 C

4.A 10.0 F capacitor has an initial charge of 100.0 C. If a resistance of 50.0 is connected across it, what is the initial current through the resistor?
1 A

5.13.3 F capacitor is charged to a potential of 55.0 V and then discharged through a 75.0 resistor.

(a) How long after discharge begins does it take for the capacitor to lose 90.0% of the following?

(i) its initial charge
1 s

(ii) its initial energy
2 s

(b) What is the current through the resistor at both times in part (a)?

(i) at tcharge
3 A

(ii) at tenergy
4 A

Explanation / Answer

1. v =120 v ,

R40 = V^2/P1 = (120)^2/(40) = 360 ohms

R60= V^2/P2 =(120)^2/(60) = 240 ohms

When connected in series, the total resistance

RT= R40+R60=360+240= 500 ohms

Current

I = 120/500 = 0.24 A

hence

P = I2 . R

P40 = 0.242 x 360 = 20.736 W

P60 = 0.242 x 240 =13.834 W

2. Total unit = 1300 W x 9h = 11.7 KWh

charge = 11.7 KWh x $ 0.13 /KWh = $ 1.52

3. Capacitance

c = 18 uF

Resistance = 100 ohm

Time costant

T = R*C = 18 x 10 -6 X 100 =18 x 10 -4

Q = Q0 e -(t/T) = V*C e -(t/T) = 6 x 18 x 10 -6 X e -(0.001/18 x 10 ^-4) = 62 uC

4.V = Q/C = 100/10 = 10 V

I =V/R = 10/100 = 0.1 A

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