Determining the power delivered by a battery -image included

[Josh225]

Homework Statement

See attachment for the question and the picture

Homework Equations

Pe= E x Is
Is = V/R

The Attempt at a Solution

I figured that in order to find the power delivered by the source, I would have to use the equation: Pe= E x Is.
E is already given to me.. 60.
That means I just have to find Is. I used the formula Is= V/R and applied it to each resistor to find the current. Here is what I got:
I1= 12A
I2= 30 A
I3= 7.5 A
I4= 3 A
I5= 5 A
I added these values up to find Is and got 57.5 A. Now I apply the formula Pe= E x Is (Pe= 60 x 57.5) and got 3450 Watts for the power delivered by the battery.

However.. The correct answer is 1.56 kW

Am I using the wrong formula completely? Where did I go wrong?
Thanks!

 

[nsaspook]

Use ohms law to check your resistance to current for voltage calculations.

 

[The Electrician]

Homework Statement

See attachment for the question and the picture

Homework Equations

Pe= E x Is
Is = V/R

The Attempt at a Solution

I figured that in order to find the power delivered by the source, I would have to use the equation: Pe= E x Is.
E is already given to me.. 60.
That means I just have to find Is. I used the formula Is= V/R and applied it to each resistor to find the current. Here is what I got:
I1= 12A
I2= 30 A
I3= 7.5 A
I4= 3 A
I5= 5 A
I added these values up to find Is and got 57.5 A. Now I apply the formula Pe= E x Is (Pe= 60 x 57.5) and got 3450 Watts for the power delivered by the battery.

However.. The correct answer is 1.56 kW

Am I using the wrong formula completely? Where did I go wrong?
Thanks!

You are calculating the current through the 2 and 8 ohm resistors as though they were each separately across the 60V battery. Actually the 2 and 8 ohm resistors are in series and the two of them could be replaced by a single 10 ohm resistor across the battery.

 

[Josh225]

Use ohms law to check your resistance to current for voltage calculations.

I checked the math, and everything checked out. Am I going about the problem in the right way?

 

[Josh225]

You are calculating the current through the 2 and 8 ohm resistors as though they were each separately across the 60V battery. Actually the 2 and 8 ohm resistors are in series and the two of them could be replaced by a single 10 ohm resistor across the battery.

I tried replacing the 8 and 2 ohms with a 10 ohm resistor. After I added them up, I got Is= 23. I threw that into the formula.. 60 (23) and got 1380 watts.
Still didn't work out.

Is that what you meant to do?

 

[The Electrician]

Current in:

5 ohm = 12A
10 ohm = 6A
20 ohm = 3A
12 ohm = 5A

12+6+3+5 = 26A

60V*26A = 1560 watts

 

[Josh225]

Current in:

5 ohm = 12A
10 ohm = 6A
20 ohm = 3A
12 ohm = 5A

12+6+3+5 = 26A

60V*26A = 1560 watts

I see now.. But how do you know to combine the 2 and 8 ohm resistors? Arent the 5 ohm and 2 ohm resistors in series?

 

[The Electrician]

Perhaps you are thinking that the wire btween the 2 and 8 ohm resistors is connected to the wire between the + terminal of the battery and the top of the 20 ohm resistor. The 2 and 8 ohm resistors are in series only and the wire between them doesn't connect to anything else.

What 1 ohm resistor are you talking about? I don't see any such resistor.

 

[Josh225]

Perhaps you are thinking that the wire btween the 2 and 8 ohm resistors is connected to the wire between the + terminal of the battery and the top of the 20 ohm resistor. The 2 and 8 ohm resistors are in series only and the wire between them doesn't connect to anything else.

What 1 ohm resistor are you talking about? I don't see any such resistor.

I apologize, I meant the 5 ohm resistor.

 

[The Electrician]

The left end of the 5 ohm resistor is connected to ground; the negative terminal of the battery is also connected to ground. Therefore the 5 ohm resistor is connected across the battery. The 2 and 8 ohm resistor are connected in series across the battery, but with respect to the battery, the 5 and 2 ohm resistors are not connected in series.

 

[Josh225]

The left end of the 5 ohm resistor is connected to ground; the negative terminal of the battery is also connected to ground. Therefore the 5 ohm resistor is connected across the battery. The 2 and 8 ohm resistor are connected in series across the battery, but with respect to the battery, the 5 and 2 ohm resistors are not connected in series.

Aaaahhhh.. Makes sense now. Thanks a lot!

 

[The Electrician]

This circuit is shown in such a way as to trick you. I usually helps to redraw the circuit. Then you can more clearly see how things are hooked up:

 

[Josh225]

This circuit is shown in such a way as to trick you. I usually helps to redraw the circuit. Then you can more clearly see how things are hooked up:

View attachment 102851

Alrighty! Ill keep that in mind. It's much easier to see that way.