Thank you for the help :)
A certain sample carries a current of 4A when the potential difference is 2V and a current of 10A when the potential difference is 4V. This samples is:
a) obeys ohm's law
b) has a resistance of 0.5 ohms at 1V
c) has a resistance of 2.5 ohms at 1V
d) has a resistance of 2.5ohms at 2V
e) does not have a resistance
I treated each pathway separately so I have
R= V/I = (2/4)= .5
R= V/I= (4/10) = .4
The answer says that it is B. I am not quiet seeing how they get that answer.
An certain resistor dissipates electrical energy at a rate of 0.5 W when connected to a 3V potential difference. When connected to a 1V potential difference, the rate of electrical dissipation:
a).5 W
B) .167 W
C) 1.5W
D) .056W
E) none of these
I am little unsure about how to go about this problem. The main equation I got was P= V^2/R. The question seems to only want power.
Pm the first, there is no correct answer.
On the second, Power= v^2/resistance so as voltage decreases by 1/3, power decreases by 1/9. One-ninth of .5 watts is ...
.056 watts, so the answer is D.
Hey there! Let me try and help you out with these questions.
For the first question, let's look at the given information. When the potential difference is 2V, the current is 4A and when the potential difference is 4V, the current is 10A. We can use Ohm's Law to find the resistance.
Using the formula R = V/I, we get R = 2/4 = 0.5 ohms when the potential difference is 2V. Similarly, R = 4/10 = 0.4 ohms when the potential difference is 4V.
Now, we need to check which option matches the resistance at 1V. Since the resistance values we calculated are for different potential differences, we cannot directly use them. However, we can assume that the resistance remains constant.
At 1V, we can assume that the resistance will be the same as when the potential difference is 2V. So, the correct answer would be option B, which states that the sample has a resistance of 0.5 ohms at 1V.
Now, let's move on to the second question.
You're absolutely right in using the equation P = V^2/R to find the power dissipation. We need to find the rate of electrical dissipation when connected to a 1V potential difference after knowing the dissipation rate of 0.5W when connected to 3V.
Let's consider the power dissipation when connected to 3V as P1 and the power dissipation when connected to 1V as P2.
Using the equation P = V^2/R, we can write P1 = (3^2)/R and P2 = (1^2)/R.
Now, let's compare the two power dissipation values. P2 is asked with respect to P1, so we can write P2/P1 = (1^2)/(3^2).
Simplifying this gives us P2/P1 = 1/9.
Since P2 is asked for, we know that P2 is equal to (1/9) * P1 = (1/9) * 0.5W.
Calculating this, we find that P2 is approximately 0.056W. So, the correct answer is option D, which states that the rate of electrical dissipation is 0.056W.
Hope that clarifies things for you! If you have any more questions, feel free to ask.
To solve the first problem:
First, let's calculate the resistance of the sample using the formula R = V/I.
For the first potential difference of 2V and current of 4A:
R1 = 2V / 4A = 0.5 ohms
For the second potential difference of 4V and current of 10A:
R2 = 4V / 10A = 0.4 ohms
Now we can compare the calculated resistance values to the given options:
a) The sample obeys Ohm's law - Since the resistance is not constant for both potential differences, this option is incorrect.
b) The sample has a resistance of 0.5 ohms at 1V - The resistance at 1V is not given, so this option is incorrect.
c) The sample has a resistance of 2.5 ohms at 1V - This option is incorrect because the calculated resistance values do not match.
d) The sample has a resistance of 2.5 ohms at 2V - This option is incorrect because the calculated resistance values do not match.
e) The sample does not have a resistance - Since we have calculated resistance values, it is clear that the sample does have resistance. Therefore, this option is incorrect.
Thus, none of the given options accurately describe the sample. The correct answer is not provided.
Now let's move on to the second problem:
Given:
Power dissipated at 3V potential difference = 0.5W
We can use the formula P = V^2 / R to solve for the resistance of the resistor.
At 3V:
P = (3V)^2 / R
Now we need to find the resistance at 1V. We know that power decreases by 1/9 (1/3 squared) based on the formula P = V^2 / R.
At 1V:
P' = (1V)^2 / R'
We can now find the new power dissipated, P', in terms of P:
P' = P * (1/9) = 0.5W * (1/9) = 0.056W
So, when connected to a 1V potential difference, the rate of electrical dissipation is 0.056W.
Among the given options:
a) 0.5W - This is the power dissipated at 3V, not at 1V.
b) 0.167W - This is not the correct value for the rate of electrical dissipation at 1V.
c) 1.5W - This is not the correct value for the rate of electrical dissipation at 1V.
d) 0.056W - This is the correct value for the rate of electrical dissipation at 1V.
e) None of these - The correct answer is d) 0.056W.
I hope that clarifies both problems for you!
To determine the answer to the first question, we can use Ohm's Law, which states that the resistance (R) of a material is equal to the potential difference (V) across it divided by the current (I) passing through it.
First, let's calculate the resistance using the given values:
For the first set of data, V = 2V and I = 4A, so R = V/I = 2V/4A = 0.5 ohms.
For the second set of data, V = 4V and I = 10A, so R = V/I = 4V/10A = 0.4 ohms.
Now, let's examine the options:
a) Obeys Ohm's Law: This is true since the material's resistance can be determined using Ohm's Law.
b) Has a resistance of 0.5 ohms at 1V: This is not correct because the resistance at 1V is not directly given and cannot be determined from the given data.
c) Has a resistance of 2.5 ohms at 1V: This is not correct because the resistance at 1V is not directly given and cannot be determined from the given data.
d) Has a resistance of 2.5 ohms at 2V: This is not correct because the resistance at 2V is not directly given and cannot be determined from the given data.
e) Does not have a resistance: This is not correct because the material does have resistance, as shown by the calculations.
Therefore, the correct answer is b) Has a resistance of 0.5 ohms at 1V since that is the resistance calculated using the given data.
For the second question, you are on the right track by using the equation P = V^2/R, where P is power, V is the potential difference, and R is resistance.
Given that the resistor dissipates electrical energy at a rate of 0.5W when connected to a 3V potential difference, we can calculate the resistance using the formula: P = V^2/R.
Plugging in the values, we have P = 0.5W and V = 3V. We can solve for R:
0.5W = (3V)^2 / R
Rearranging the equation, we get:
R = (3V)^2 / 0.5W
= 9V^2 / 0.5W
= 18V^2/W
Now, to determine the power dissipation when connected to a 1V potential difference, we can substitute V = 1V into the equation:
Power = V^2 / R
= (1V)^2 / (18V^2/W)
= 1/18 W
Therefore, the correct answer is D) 0.056W since the power dissipation when connected to a 1V potential difference is 1/18 W.