How Does Closing a Switch Affect Voltmeter and Ammeter Readings in a Circuit?

[clope023]

Homework Statement

The circuit in the attached figure (excuse the terrible drawing, I suck at using the paint feature) all meters are idealized and the batteries have no appreciable internal resistance

a) Find the reading of the voltmeter with the switch S open

b) With the switch closed, find the reading of the voltmeter and the ammeter. Which way does the current flow through the switch?

Homework Equations

V = I/R (ohm's law)

$$\Sigma$$I = 0 (junction law)

$$\Sigma$$V = 0 (loop law)

The Attempt at a Solution

To find the reading of the voltmeter I attempted to find the currents going through the 2 emf's and then using the junction rule to find current at point a (the top) and then using ohm's law with that current and the 75 ohm resistor to find the reading of the voltmeter, with the switch open I treated that segment as if the switch line wasn't there, here's what I got

for loop 1 (with the 25V emf)

25V-I1(100 ohms) = 0

25V = I1(100ohms), I1 = .25A

for loop 2 (with the 15V emf)

15V-I2(75 ohms) = 0

15V = I2(75 ohms), I2 = .20A

junction rule .25A-.20A-I = 0, I = .05A

ohm's law for voltmeter

I = V/R, V = IR -> V = (.05A)(75ohms) = 3.75V

I'm not sure if I did the question correctly, if anyone can check it for me I would appreciate it, thank you.

 

[dynamicsolo]

It looks like your attachment is coming through finally, so we can look at the circuit.

With the switch open, there is only one loop containing two batteries in series and two resistors in series. That means there's only one current to solve for in part (a).

The ammeter is on the open branch, so naturally it will read nothing. The voltmeter is measuring the potential "drop" across the 15-V battery and the 75-ohm resistor. So you'll need to find the voltage drop across the resistor, then "add in" the 15 volts from the battery appropriately.

After that, we'll deal with part (b).

 

[Moetasim]

Your approach to finding the reading of the voltmeter is correct. However, you should also consider the direction of the current in each loop. Since the 25V emf is in the opposite direction as the 15V emf, the resulting current in loop 1 should be negative (-0.05A) and the current in loop 2 should be positive (0.20A). This will result in a net current of 0.15A at point a, which will give a reading of 11.25V on the voltmeter.

With the switch closed, the current will flow from the positive terminal of the 15V battery, through the ammeter, through the switch, and then through the 75 ohm resistor, back to the negative terminal of the 15V battery. This will result in a current of 0.20A through the ammeter and a reading of 15V on the voltmeter. The direction of the current through the switch will be from the positive terminal to the negative terminal of the battery, as indicated by the arrow in the circuit diagram.