Voltmeter Question

Thread Starter

xz4chx

Joined Sep 17, 2012
71
A voltmeter will be used to measure Va. Voltmeter resistance is 100 Ohms and full scale voltage 20 V

What does the Voltmeter read at Va?

I can use KVL, KCL, series and parallel rules for resistors, voltmeters rules and voltage and current divider rules



1. Is that do I need to simply the bottom section of the circuit in order to find Va? To me it seems like I don't need it but I want to be 100%.

2. If I turn the 15 Ohm and 22 Ohm resistors at the top into 37 Ohms (In series), can I add in the 33 Ohm resistor into that or do I need it in order to find Vb (which I am assuming I need for the dependent source.)? Because If I took out the 33 Ohm resistor in series then Vb would not be valid anymore, Correct?

3. What do I need in order to find Vb if I am going in the wrong track on question 1,2?

4. If I break the bottom section of this circuit down into the dependent current source connected to a resistor and then used the 7V voltage source and the 33 ohm resistor, would this be the best strategy to find Vb. (I know this question 1 invalid, but I just thought of it after all of this.
 

WBahn

Joined Mar 31, 2012
33,142
A voltmeter will be used to measure Va. Voltmeter resistance is 100 Ohms and full scale voltage 20 V

What does the Voltmeter read at Va?

I can use KVL, KCL, series and parallel rules for resistors, voltmeters rules and voltage and current divider rules



1. Is that do I need to simply the bottom section of the circuit in order to find Va? To me it seems like I don't need it but I want to be 100%.
Correct - you can ignore the bottom if all you want is Va, since Vb does not depend on anything in the lower part. Now, if you were trying to find Vc, you could NOT ignore the top part, because Vb does depend on the top part.

2. If I turn the 15 Ohm and 22 Ohm resistors at the top into 37 Ohms (In series), can I add in the 33 Ohm resistor into that or do I need it in order to find Vb (which I am assuming I need for the dependent source.)? Because If I took out the 33 Ohm resistor in series then Vb would not be valid anymore, Correct?
Correct as far as needing to keep the 33Ω resistor separate so that you can determine Vb, but you can't combined the 15Ω and the 22Ω resistors because, with the meter attached, they are not in series. Isn't the goal to find what the meter reads?

3. What do I need in order to find Vb if I am going in the wrong track on question 1,2?
You need to add in the meter resistance and, even then, you can't combine the resistors that Va is across with other resistors because THAT is the voltage you are trying to calculate. Of course, you can combine them to perform an intermediate step, such as finding the current, and then expand them back out to find Va given that current.

4. If I break the bottom section of this circuit down into the dependent current source connected to a resistor and then used the 7V voltage source and the 33 ohm resistor, would this be the best strategy to find Vb. (I know this question 1 invalid, but I just thought of it after all of this.
You're right -- it's invalid. So I have no idea how to answer it other than, no, it wouldn't be the best strategy to find Vb.

Consider these two questions separately.

Q1) Given a value of Vb, what is the current flowing in the upper loop?

Q2) Given a value of current flowing in the upper loop, what if Vb?

Doesn't this yield two equations in two unknowns?
 

Thread Starter

xz4chx

Joined Sep 17, 2012
71
So I first need to know what the value of Va is without the voltmeter attached as i have to calculate percent error of tested and accepted reading. So to do that I used

Making \( i_1 \) the current through the 33 ohm resistor heading toward \(V_a\)

70i1 I got from adding the series of resistors since it is not attached to the voltmeter.
\( -3v_B + 70i_1 = 0
33i_1 + v_B + 7 = 0 \)

Then Solve for \(v_B\) and \( i_1\) and then use those values to solve \(v_A\)

After that I attach the 100 ohm 20 V full scale voltmeter (modeling as a resistor) to the circuit at \(V_a\) I can't assume the voltmeter is 20V right? If not
Put the two parallel circuits together and then use those to connect the rest in series and solve for v_B then expand that out to solve to v_A
 
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