A battery of 10.0 V is connected across a resistor of 40.0. Calculate the power delivered to the resistor.?

Question

A. 400 W

B. 10 W

C. 2.5 W

D. .25 W

Bulb P is rated 60 W, 110 V and Bulb Q is rated 100 W, 110 V. Which of the bulbs has the higher resistance?

A. They have the same.

B. Cannot be determined.

C. Bulb P

D. Bulb Q

Thanks guys! I would really appreciate it if someone could give me a quick explanation of how to figure this out. I'm sure it's quite easy, I just can't find it in my book anywhere. The whole book itself is very vague. But, if not, that's fine. Thanks anyways!

Answer 1

1) Power law:  Power = voltage^2 / resistance

P = 10^2 / 40

P = 100 / 40 = 2.5W

2) As a simple resistance problem bulb P has the higher resistance.

Again, use the power law:

60 = 110^2 / resistance

resistance = 110^2 / 60 = 201.6 Ohms

100 = 110^2 / resistance

resistance = 110^2 / 100 = 121 Ohms

(In the real world, filament light bulbs are a bit more complex, but the answer is still the same for this example)

.

Answer 2

Assuming no other losses, e.g. in the wire....

Power = Volts^2 divided by resistance

i.e. 10 x 10/40 = 2.5 Watts

So you see lower resistance is higher the power.  Now you can answer Question 2.

Answer 3

1.  answer (c)  power = V^2/ R  so 100/40 = 2.5W

2. answer(d)  the higher the wattage the higher resistance.