- #1

brentwoodbc

- 62

- 0

## Homework Statement

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.

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- Thread starter brentwoodbc
- Start date

- #1

brentwoodbc

- 62

- 0

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

tyco05

- 161

- 0

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

brentwoodbc

- 62

- 0

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

then I found voltage

current

v=ir

14=Ix16.67

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

- #4

berkeman

Mentor

- 64,121

- 15,319

- #5

brentwoodbc

- 62

- 0

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

berkeman

Mentor

- 64,121

- 15,319

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

i1+i2+i3...=0

v1+v2+v3=0

Im stuck.

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.

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