Can somebody check my answers (DC-circuit)

[cseet]

Hi all

need somebody to go thru my answers are actually correct for DC circuits...

thanks in advance...

question:
Two 1.55 V batteries are connected in series to power a lamp. One battery has an internal resistance r1 = 0.208 ohm, and the other has r2 = 0.169 ohm. When the switch is closed a current of 0.55 A flows and the lamp lights up.

answer:
What is the terminal voltage of the battery combination and what is the resistance R of the lamp filament?

question:
What fraction of power is dissipated in the batteries?
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can somebody advise me on how I get the fraction of power answer?
my conclusion was (correct me if I was wrong):

Power = V I, so

power in internal r Vi I
------------------------ = ------ = am I in the right track?
power in whole circuit V I

thanks again
cseet

 

[Doc Al]

can somebody advise me on how I get the fraction of power answer?
my conclusion was (correct me if I was wrong):

The total power delivered by the batteries is given by P = VI, where V is the total voltage of the two batteries. This power is dissipated in the resistance of the circuit, some of which is internal to the batteries. The power dissipated in a resistor is P = V_RI, where V_R is the voltage drop across the resistor. From Ohm's law, the voltage drop across a resistor is V = IR, so the power dissipated in a reistor becomes: P = I²R.

So... Find the power dissipated in the internal resistance of the batteries and divide by the total power. So you are on the right track if by V_i you meant the voltage drop due to the internal resistance.

 

[Chen]

By the way, if I'm not mistaken the fraction of power that is dissipated outside the batteries is called the circuit's efficiency (I think it's denoted by the letter \mu).