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I have an open circuit that looks like this:

I can measure the voltage at these points using a voltmeter:

A -> D: 4.65V (the batteries)

B -> D: 4.3V

C -> D: 4.23V

Say I did not have a voltmeter: What formula would I use to determine the voltage between B -> D and C -> D?

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Without a voltmeter you would have

\$V_{cd} = V_{bd} = V_{ad} \approx 4.65 \mathrm{V}\$.

The reason is that with an open circuit, the current through each resistor is 0, and so by Ohm's law, the voltage drop across each resistor is 0.

The only reason you don't read this result with your meter is that there is a small leakage current through the voltmeter when you connect it to the circuit.

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Adding to what Allan & The Photon said:

Your results are in good agreement with what you would expect if your meter has a resistance of close to 1 megohm. The results are within the tolerance range expected due to typical component value variations.

I = V/R
Across the 82 k resistor you see (4.65 - 4.3) = 0.35V
The current that causes this can be determined from
I = V/R = 0.35/82k = 4.268 microamps.
When measuring from point B to ground you would have that current flowing through Rmeter.
R = V/I = 4.3 V /4.268 uA = 1.007 megOhm

Similarly, when measuring from point C to ground similar reasoning indicates
Rmeter = (4.23 / (4.65-4.23)) x (82k + 18k) = 1.007 megohm (again).
I have purposefully not spelt out the steps in arriving at that equation - it makes the same assumptions as in the prior calculation.

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The only reason you measure a lower voltage than the battery voltage on the B->D and C->D cases is that your voltage measurement device has some resistance inside that allows some current to flow.

If you so not inject the volt meter into the circuit because you did not have one then the B->D and C->D voltage levels would be the same as the A->D measurement. No current across the 82K and 18K resistors means no voltage drop across them.

Since every measurement device will have some resistance (some as low as the K ohm range whilst others up into the multi megohms range) you can place an equivalent meter resistance in the circuit and use KCL to solve the circuit for voltage drops.

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If assumption is that the meter has a 1 mega ohm resistance then the total resistance between ground and the 82K resistor is: 1,820,000 ohms. The current would not include the 18K ohm resistor in this case, therefore the total current would be: 4.65 volts / 1,820,000 ohms = 2.555 micro Amps The voltage drop across the 82K ohm resistor would be: 2.555 micro Amps X 820,000 ohms = 2.095 Volts. The voltage drop across the internal 1 mega ohm resistor of the meter would be: 2.555 micro amps X 1,000,000 = 2.555 volts. Total Voltage: 2.095 + 2.555 = 4.65 Volts.

If you take the voltage measurement from C to D (C to ground) it now includes both resistors and the 1 mega ohm resistor of the meter to get total resistance. Divide total voltage by total resistance to get the total current and use that to compute the voltage across each resistor. This is all based on the meter resistance included in the circuit. Otherwise everywhere in the circuit is just an open circuit with 4.65 Volts when the meter is not included in the circuit since open circuit equals 0 current (amps).

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