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Homework Help: Can somebody check my answers (DC-circuit)

  1. Mar 28, 2004 #1
    Hi all

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

    thanks in advance...

    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.

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

    What fraction of power is dissipated in the batteries?

    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
  2. jcsd
  3. Mar 28, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    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 = VRI, where VR 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 Vi you meant the voltage drop due to the internal resistance.
  4. Mar 28, 2004 #3
    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 [tex]\mu[/tex]).
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