1)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
I got C. b/c of higher potential difference.
2)A battery of 10.0 V is connected across a resistor of 40.0. Calculate the power delivered to the resistor.
A)400 W
B)10 W
C)2.5 W
D)25 W
10^2/40.0
I got A
On the first, potential difference has nothing to do with it.
Resistance=voltage ^2/powerrating
On the second, no
Power= 10^2/40. That is not 400
I meant C on 2.
For question 1, the correct answer is C) bulb P. This is because the resistance of a bulb is inversely proportional to the power it consumes. Therefore, the bulb with a lower power rating (60W) will have a higher resistance compared to the bulb with a higher power rating (100W), assuming they have the same voltage.
For question 2, the correct answer is C) 2.5 W. The power delivered to a resistor can be calculated using the formula P = V^2/R, where P is the power, V is the voltage, and R is the resistance. Plugging in the given values, we get P = (10.0V)^2 / 40.0Ω = 2.5W.
1) To determine which bulb has the higher resistance, we can use Ohm's law, which states that resistance (R) is equal to voltage (V) divided by current (I), R = V/I. Since both bulbs have the same voltage (110 V), we can compare their power ratings (P) to determine the current flowing through each bulb. The formula P = VI can be rearranged to find current, I = P/V.
For bulb P, I = 60 W / 110 V = approximately 0.545 A.
For bulb Q, I = 100 W / 110 V = approximately 0.909 A.
Since resistance is inversely proportional to current (R = V/I), we can see that bulb P has a higher resistance than bulb Q because it has less current flowing through it. Therefore, the correct answer is option C) bulb P.
2) To calculate the power delivered to the resistor, we can use the formula P = V^2/R. Here, V is the voltage across the resistor and R is the resistance value.
Given V = 10.0 V and R = 40.0 Ω, we can substitute these values into the formula:
P = (10.0 V)^2 / 40.0 Ω = 100.0 V^2 / 40.0 Ω = 2.5 W.
Therefore, the power delivered to the resistor is 2.5 watts. The correct answer is option C) 2.5 W.