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 Quote by chroot You need to find the current through the total circuit, which is composed of both the external and internal resistances. Next, you need to find the voltage drop across the internal resistance. You can use Ohm's law. Finally, you know how much voltage is being "lost" on the internal resistance, so you know what voltage will appear on the battery's terminals. - Warren
First of all, I'd like to thank you for your quick reply. However, I'm not quite sure if I understand you correctly.

Ok, so the battery's emf is 6V, internal resistance is 0.6 Ohms, and the circuit's net resistance is 7.20 Ohms.

When you said:

 Quote by chroot You need to find the current through the total circuit, which is composed of both the external and internal resistances.
I took that as...the total circuit has a resistance of 7.8 Ohms.
So, I have a current of 6.0/7.8 = 0.769 Amps ??

 Quote by chroot you need to find the voltage drop across the internal resistance. You can use Ohm's law.
So...I found that as Current X Internal Resistance = 0.462 V

6.0 V - 0.462 V = 5.54 V, which is the answer.

So...I don't need to worry about the resistance through the circuit?