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I am given the following circuit and told to calculate the voltage measured by the voltage meter:

enter image description here

My first thought was to calculate the voltage through \$R_k\$ (at the red dot) and the voltage through \$R_i\$ (at the blue dot) and have the result be their difference. Is my train of thought correct? If not how do I calculate said voltage?

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4 Answers 4

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Voltage through a resistor?...wrong thinking.

Currents flow through resistor, and thinking of current paths is the proper approach to this problem.

The source of current is given as \$I_o\$, shown as \$I_{in}\$ in the diagram below. This current \$I_{in}\$divides between two paths:

  • One path flows through Rx
  • Other path flows through Rk, Ri, Rk
  • Assume that no part of \$I_{in}\$ current flows through the meter "u".

re-drawn OP schematic showing current paths.


You should already know how to calculate the fraction of current that flows through one path versus the other path, when resistances of both paths are known.
Once you know the fraction of \$I_{in}\$ that flows through \$R_i\$, voltage across \$R_i\$ is known.

One should also be aware that as a practical matter, there exists no voltage-measuring device that can measure voltage without pulling a little current. Some tiny fraction of \$I_{in}\$ will take the voltmeter path. Some voltmeters are almost ideal, meaning that the current through them is extremely small.
In this simple example, we assume that the voltmeter is "ideal" (\$I_{meter}=0\$)

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  • \$\begingroup\$ So if \$I_1\$ is the fraction of the current flowing through \$R_i\$ with \$I_1 = I_0 \cdot \left(\dfrac{R_x}{2R_k+R_i}\right)\$ then \$V_i = I_1 \cdot R_i\$? Or do I have to account for both \$R_k\$ in the last step aswell? \$\endgroup\$
    – Tahunie
    Commented Aug 24, 2022 at 19:13
  • \$\begingroup\$ Your simple idea for \$V_i\$ is a correct solution. There will be voltage drop across the two \$R_k\$, so the voltage measured from blue_dot to red_dot will be less than the voltage drop across \$R_x\$. All we need for a solution is value of \$R_i\$ and the current through it (\$I_1\$) \$\endgroup\$
    – glen_geek
    Commented Aug 25, 2022 at 0:12
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There is a question of terminology here. You say you are tasked with finding the voltage measured by the 'voltage meter'. In the drawing there is a current source, a resistance \$R_x\$ and a box labeled 'Voltage Meter' with resistors \$R_i\$ and \$R_k\$ and a meter movement labeled '\$\mu\$' or 'u'.

If we take this to mean that the voltage meter is made up of the movement and resistors in the box, the voltage that it will measure is the voltage across \$R_x\$.

This voltage will be \$I\$ times the total resistance, \$R_x\$ in parallel with the voltage meter resistance.

The problem is how to determine the voltage meter resistance. If we take the \$R_k\$ values as the connection resistance (and assume they are equal) and \$R_i\$ as the meter internal resistance it would be $$2R_k + R_i$$ If the meter movement has resistance that is not included in \$R_i\$ then that has to be taken into account and it becomes $$2R_k + (R_i || R_{movement})$$

Since no value is given for the meter movement's resistance you would either have to assume that it is part of \$R_i\$, that it is large enough to not matter, or that you can't answer without knowing it. I would lean towards it being part of \$R_i\$.

If on the other hand, the 'voltage meter' you are supposed to find the measurement of is not the boxed area but the meter movement \$\mu\$, then look at the other answers here.

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  1. Calculate the current through the \$R_k + R_i + R_k\$ path using the current divider rule.

  2. The voltage across \$R_i\$ (which is what is measured by the voltmeter) is given by Ohm's law.

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The parallel resistance Rq of Rx and the series combination of Rk+Ri+Rk can be calculated like two parallel resistors. Rq = R1*R2/(R1+R2 or, equivalently, 1/(1/R1+1/R2)

The voltage across the combination is just Io*Rq

Then you have a voltage divider like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Which is just a voltage divider- and you can combine the two Rk resistors.

So V = Ri/(Ri+2*Rk)

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