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Two Circuits connected in parallell

Hi guys

In a bit of a pickle with this question i have,

Question: "Two circuits, A and B, are connected in parallel to a 25v battery, which has an internal resistance of 0.25 Ohms. Circuit A consists of two resistors, 6 Ohms and 4 Ohms, connected in series. Circuit B consists of two resistors, 10 Ohms and 5 Ohms, Connect in series. Determine the current flowing in and the voltage drop across each of the four resistors. Also find the power dissipated by the external circuit.

Just wondering you someone could help me with this?

I think i have interpreted the question right and drawn what i think the circuit should be.

Cheers

Question 4 Circuit.jpg11.2KB

Reply 1

Stonebridge

The circuit is ok.
So do you know how to add resistors in series and parallel?

Reply 2

Ramjam

Original post by Stonebridge

The circuit is ok.
So do you know how to add resistors in series and parallel?

Yeh I understand that, but wouldn't adding them and making one circuit prevent me from working out the voltage drop across each one?

Reply 3

Stonebridge

Original post by Ramjam

Yeh I understand that, but wouldn't adding them and making one circuit prevent me from working out the voltage drop across each one?

No.
Making a single circuit (one combined resistance) enables you to find the current being delivered by the battery, the main current in the circuit. E=IR where R is total resistance and E = emf of battery.

You have that main current which then splits between two parallel branches. One is 10 Ohm and the other 15 Ohm.
One you have the current in the two parallel branches, V=IR gives you the pd across each resistor.

Reply 4

Ramjam

Original post by Stonebridge

if work it all okay can you check my workings?

Reply 5

Stonebridge

Yes. Post them here and we can take a look.

Reply 6

Ramjam

Original post by Stonebridge

Yes. Post them here and we can take a look.

Hi sorry it took so long, been sat under my car all day, the downside of owning a classic car! :frown:

Heres my workings, think they are right.

All OK to the end of the 1st page.
Then
The pd across the parallel branches is not 25V
There is some lost on the internal resistance. Subtract this from 25V to find the correct value for the next part.
After that the method is correct for finding the pds across the individual resistors.

Reply 8

Ramjam

Original post by Stonebridge

Brilliant cheers , thanks for the feedback

Reply 9

Ramjam

Original post by Stonebridge

Do you mean work out the drop for the internal resistance first, so

"total Current (4A) * 0.25 = 1v"

So the supply is in fact 24v now work out the drop over each component?

Reply 10

Stonebridge

Original post by Ramjam

Do you mean work out the drop for the internal resistance first, so

"total Current (4A) * 0.25 = 1v"

So the supply is in fact 24v now work out the drop over each component?

Yes. The pd across the external circuit (the parallel branches) is 24V
Use that, and not 25V, and you will get the correct answers with the method you have there.

One point.
The power dissipated in the external circuit is 24V x 4A
The power generated by the battery is 25V x 4A
You can probably tell me where the missing power went.