Power EK. 156

[SaintJude]

View attachment 18706

Answer is here, hidden in white: I

1.) To find the power, Ek used P = i^2 (R). Why didn't they use P=iV? Is it because the former equation is applicable only when voltage is constant?

2.) I though the current would be same across each resistor in II, so the same power would be formed as I. That's wrong. Why?

3.) Assume, I had known to rank the effective resistances. The power would be brighest to dimmest in the rank: I, III, II. Yes?

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[MedPR]

View attachment 18706

Answer is here, hidden in white: I

1.) To find the power, Ek used P = i^2 (R). Why didn't they use P=iV? Is it because the former equation is applicable only when voltage is constant?

2.) I though the current would be same across each resistor in II, so the same power would be formed as I. That's wrong. Why?

3.) Assume, I had known to rank the effective resistances. The power would be brighest to dimmest in the rank: I, III, II. Yes?

1. They use P=I^2R because when you find the Req of the circuits, two of them are calculated by adding them in series (the first one has only 1 resistor, so I don't think it matters). And yes, since the batteries are of different size, the voltage is going to be different.

2. The current is identical across both resistors in circuit II, but the equivalent resistance will be different than in circuit I, assuming all resistors in all three circuits are equivalent. For instance, if all the resistors are 3ohms, circuit 1 has Req of 3, circuit 2 has Req of 6, etc.

3. Yup.

 

[milski]

View attachment 18706

Answer is here, hidden in white: I

1.) To find the power, Ek used P = i^2 (R). Why didn't they use P=iV? Is it because the former equation is applicable only when voltage is constant?

Both are equivalent, use whichever one is easiest at the time. P=I^2 * R = U/R*I*R=IU

2.) I though the current would be same across each resistor in II, so the same power would be formed as I. That's wrong. Why?

The current is the same in each of them. It's different than the current in the first circuit. Should be I/2, if all bulbs are the same and I is the current in the first circuit. That makes the power in the second circuit I^2R/4+I^2R/4=I^2R/2

3.) Assume, I had known to rank the effective resistances. The power would be brighest to dimmest in the rank: I, III, II. Yes?

Equivalent resistance for the last circuit is R+1/(1/R+1/R)=3R/2. Current through the single bulb will be 2I/3 and I/3 for the other two. The total power is (4/9I^2+1/9I^2+1/9I^2)R=2I^2R/3

Correct, the power is I, III, II.

 

[SaintJude]

I understand that the equations are equivalent but you wrote this below a while ago. I thought you were saying each equation is applicable under certain conditions.

" 1. P=VI
2. P=I^2R
3. P=V^2/R

Equation 2 says that if you increase resistance and hold current constant, you increase power.

Equation 3 says that if you increase resistance and hold the potential constant, you decrease power.

 

[milski]

They are always applicable. What I was saying was related to using them to make a qualitative assessment of how power changes under certain conditions. For example if you increased the voltage and the resistance in some circuit while maintaining the same current you cannot infer anything about the power from P=V^2/R - you increased both but without knowing by how much you cannot say anything about the ration V^2/R. Using P=I^2R in that case would be much more useful.

I think that can be also seen in the problem where you were discussing the two resistors with different length - using P=V^2/R was the most straightforward choice to get the answer there.

 

[SaintJude]

Gottcha! You realize, you could brush your teeth tomorrow, wear some sweatpants, take the MCAT..... and get a 40.

 

[MedPR]

I understand that the equations are equivalent but you wrote this below a while ago. I thought you were saying each equation is applicable under certain conditions.

" 1. P=VI
2. P=I^2R
3. P=V^2/R

Equation 2 says that if you increase resistance and hold current constant, you increase power.

Equation 3 says that if you increase resistance and hold the potential constant, you decrease power.

Since they are equivalent equations, could could probably use them all in any situation given enough information. It's analogous to the kinematics equations. If I give you v final, v initial, and acceleration and ask you to find displacement, you could plug in the numbers to a=v/t, then plug t into x=vot+1/2at^2, or you could do it directly with v^2=vo^2 + 2ax.