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Physics Electricity Question watch

1. Hi

So basically there's a physics hmwk question I am stuck on. So there's a circuit with a battery with emf 12V, two 10 Ohm resistors in parallel, and another 10 Ohm resistor in series with the battery. The current is 0.5A.

The internal resistance of the battery is emf over current then minus the resistance, which I have worked out as 12 over 0.5 then minus 15 which equals 9 Ohms.

Then it asks the total energy supplied by the battery in 2 seconds. I know E = I2Rt but does the R (resistance) include the internal resistance of the battery?

If it doesn't then its the current squared which is 0.25 x 15 x 2 which is 7.5J.

If the internal resistance in the battery is included then the 9 Ohms is included so it's 0.25 x 24 x 2 = 12J.

Which is correct?
2. (Original post by funky386)
Hi

So basically there's a physics hmwk question I am stuck on. So there's a circuit with a battery with emf 12V, two 10 Ohm resistors in parallel, and another 10 Ohm resistor in series with the battery. The current is 0.5A.

The internal resistance of the battery is emf over current then minus the resistance, which I have worked out as 12 over 0.5 then minus 15 which equals 9 Ohms.

Then it asks the total energy supplied by the battery in 2 seconds. I know E = I2Rt but does the R (resistance) include the internal resistance of the battery?

If it doesn't then its the current squared which is 0.25 x 15 x 2 which is 7.5J.

If the internal resistance in the battery is included then the 9 Ohms is included so it's 0.25 x 24 x 2 = 12J.

Which is correct?

The current is 0.5A and the e.m.f. of the cell is 12V in which case the total energy expended is E = I2(r+R)t = 0.25(9+15)2 = 12J.

The battery terminal voltage in this instance is:

V = 0.5 x 15 = 7.5V

So the energy supplied to the load is:

Eload = I2Rt = 0.25 x 15 x 2 = 7.5J

And the energy expended as heat by the battery internal resistance is:

I2rt = 0.25 x 9 x 2 = 4.5J

In other word total energy used is 7.5J + 4.5J = 12J.

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