# Electric power question

1. Oct 22, 2005

### catalyst55

Q) Calculate the voltage at lamp 3.

i've apparently oversimplified the question with my answer of 10v.

if anyone could that could help, that'd be great.

2. Oct 22, 2005

### Chi Meson

OK

What you have is a parallel circuit with three branches.
Each branch has resistance (in series) from the lamp and the wires.
The resistance of the lamp is found by examining the power equation (at 12 volts, the lamps produce 12 watts, so what must be their resistance?).
Each branch gets the full 12 volts, but the wires in the lower branch have dropped the voltage by exactly one; what's the total resistance of those two wires there.

OK so far?

edit:
OOPS, check that!

There are three branches, but, it's a series in parallel in series in parallel:

lamp 3 is in series with two wires
this series is parallel with lamp 2.

This parallel is in series with two more wires

this series is parallel with lamp 1 and two wires.

Last edited: Oct 22, 2005
3. Oct 22, 2005

### catalyst55

Hi
The preceding question asks to find the total resistance of the connecting lead between lamp 2 and the battery.

This is easily done by modelling the resistive lead as a resistor and drawing the appropriate circuit diagram.

This gives:

R/(R+12) = 1/12, hence R = 1.1ohm

So the resistance of the leads connecting lamp 2 is 1.1ohms.

Then we're asked to find the voltage at lamp 3.

If we, again, draw the circuit diagram, and model the resistive leads as resistors, we gets this circuit, where the resistance of the resistors is 1.1ohm, and 12ohm for the lamps (right = lamp 3; left = lamp1)

In order to find the voltage across lamp3, we must know the current.

So first we'll find the total resistance (this is where my answers start to deviate from the book's):

R(T) = [ 2(1.1+12)^-1 ] ^-1 = 6.55ohm

Which gives a total current of 1.83A.

What have i done wrong? And, assuming that I'm right, how would i find how the current is split? - It's not split equally, as i would have thought.

Thanks a lot.

Last edited: Oct 22, 2005
4. Oct 22, 2005

### Chi Meson

I agree with you so far. With this current through the top-left resistor (according to your drawing), the voltage drops 2.02 volts. This leaves 9.98 volts across lamp 2 and also 9.98 volts across the lamp 3/resistor parallel.

The current though lamp 3 should be .76 amps, and the voltage should drop across the last resitor by .83 volts leaving 9.14 volts across lamp 3.

This is off from the given answer, I think, because the restance of the wires should be 1.09 ohms instead of 1.1 ohms during calculations.