1. Apr 24, 2005

Clari

For the circuit shown in the diagram,

a. Find the readings on teh ammeters A, B and C( assumed to have effectively 0 resistances)

b. Find the p.d. between X and Y.

c. Find the power dissipated as heat in the circuit.

d. Find the power delieverd by the 12 V cell.

e. Account for the differences between c. and d.

Sorry, I am not asking you to do the homework for me, but I really don't know how to start off when so many cells are connected together.

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Last edited: Apr 24, 2005
2. Apr 24, 2005

SpeedBird

maybe try finding voltages due to the individual batteries at first.
i.e. short circuit the other two.
im not sure if this is the way to go about it. i may be completely wrong!
it's a suggestion anyway :-)

3. Apr 24, 2005

OlderDan

You have three unknowns in the problem, the three currents. Once you know those, you should be able to do the rest. Call the top 8ohm Ra, the 10ohm Rb and the middle 8ohm Rc to go along with the labels on the meters. Use the same scheme for the currents Ia, Ib, Ic. Clearly the top of the 2V source is at a lower potential than the tops of the 12V source and the top of the 3V source, so a reasonable assumption is that current Ic will flow down through C. Assume Ia flows to the right and Ib to the left. It does not matter if you get these directions right. If you guess wrong, you just get negative values for the answer. Under these assumptions

Ia + Ib = Ic

Now take a walk around the left loop and add up all the changes in potentail. The sum of the changes must add up to zero. Take a walk around the right loop and do the same. Now you have three equations for the three unknown currents. Solve them and you are on your way.

4. Apr 25, 2005

Clari

Thank you for all of your help! : ))
i have worked out parts a and d of this question:

a.) Round the left loop, 12 - 8Ia - 8Ib - 2Ic - 2 - 2Ia= 0 -------> Ia + Ic = 1
Round the right loop, 3 + 10Ib - 8Ic - 2 - 2Ic + 5Ib =0------> 15Ib - 10Ic = -1
Since Ia = Ib + Ic, solving these three questions gives the values of ammeter A reading as 0.6A, ammeter B reading as 0.2A, and ammeter C reading as 0.4 A.

d.) Power dissipated by the 12V cell = I E = 0.6*12 = 7.2 W
For E is the e.m.f. of the 12 V cell

For part b, I tried to solve it like: Power = I^2R +IE = 0.4^2 * 2 + 0.4 *2 = 1.12W
Since P = IV, 1.12 = 0.4V, V = 2.8 V.....it is wrong! The answer is 6V, but i don't understand it.