# Volt/amp question

1. May 20, 2009

### brentwoodbc

1. The problem statement, all variables and given/known data
whats are the readings at the voltmeter and the ampmeter?

You can see my work, I got V=14.74? and I=0.72A? I need help.
thanks.

2. May 20, 2009

### tyco05

From what I can tell your equivalent resistances are ok.

If the supply is 12V - YOUR calculation for current gives you the current through R1.

This splits at the next junction - in inverse proportion to the resistances of R2 and R3.

3. May 20, 2009

### brentwoodbc

I made a mistake on the resistors on my paper (so thats wrong) heres what tried.

I simplified the resistors and got 20/3 ohms plus 10ohms = 16.67ohms

then I found voltage
current

v=ir
14=Ix16.67
I=0.84 amps.

But that is the circuits current how do you find it at the metre? same sort of thing is confusing me with the voltmeter.

then I went back to the volt meter with the 20 Ohm resistor so

v=IR
v=.84x20
v=16.8Volts?

4. May 20, 2009

### Staff: Mentor

I don't see simplificcation solving this circuit. Just write the 2 KCL equations and solve them simultaneously...

5. May 20, 2009

### brentwoodbc

how do you do that, so you find current and voltage and plug into

i1+i2+i3....=0
v1+v2+v3=0

Im stuck.

6. May 20, 2009

### Staff: Mentor

There are two summing nodes that I see where I would write the 2 KCL equations (sum of currents into a node must equal zero). The first node is between the two 10 Ohm resistors and the Ammeter, and the second node is on top of the final 20 Ohm resistor.

Write those two equations, and you should be able to solve the circuit. It's true that the voltage drop across an ideal ammeter is zero, so you can probably collapse the circuit and solve, but since they are asking for the I and V, my first instinct would be just to solve the full KCL, and recognize the simplification later.