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# Why the voltmeter Shows this?

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#### Electronman

##### New Member
Hi,

Why the voltmeter shows 12V while the above SW is open And was open before( I never closed it!)?

Thanks

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There is no charge on C1 therefore it is a short circuit and the multimeter is just measuring the voltage across C2 = 12V.

The Voltage across C2 is 12 V.
The voltage across C1 is 0 V.

12 + 0 = 12 volts

Does not make any sense to me!

Is C1 in a closed circuit so that we could say that at t=0 it is shorted? is it able to reach to 12v too at infinite times?

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The voltmeter in your diagram is not real. To make it real, place a 10MΩ resistor across it.

Then it will begin at 12V then decay to zero.

The voltmeter in your diagram is not real. To make it real, place a 10MΩ resistor across it.

Then it will begin at 12V then decay to zero.

What do you mean really?! If it is not ideal so why the circuit is closed at place of C1?

I read 10.5 volts by my multimeter in real when the source voltage was 12V???!

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What is t=0 charge on C1. Answer, 0vdc. It takes some load from voltmeter to charge it to -12 vdc so there will zero volts across meter.

What is t=0 charge on C1. Answer, 0vdc. It takes some load from voltmeter to charge it to -12 vdc so there will zero volts across meter.

What does this mean? What he drew is an "ideal circuit".

Most all transient simulation programs start simulation with zero volts on caps. Your battery current would be near infinity at t=0 to charge the bottom cap. Some sim programs may abort with error for this.

If you want to talk real world, for a 10 meg ohm DVM, it would take over 50 hours for the DVM to get from 0 vdc to a reading of 10 vdc once battery was inserted. (this is assuming 100000u means 0.1 farad).

You have also discovered why you never want to leave a CMOS input of an I.C. open.

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Mathematics is a wonderful tool to understand the complexity of the electrical behavior of a capacitor. The voltage across the capacitor is governed by V = Q/C where Q is the charge and C is the capacitance of the capacitor. The capacitor has not been charged. As a result, the charge on the capacitor is approximately equal to zero coulombs and by the definition of the voltage equation given above V = 0/C = 0. Hence, the voltage across the terminals of the capacitor is zero volts. The rest can be understood by applying Ohms Law (V = I *R). The internal resistance of a CAPACITOR can be modeled as a short when initially connected to a power source. I hope this assists you understand your question.

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Mathematics is a wonderful tool to understand the complexity of the electrical behavior of a capacitor. The voltage across the capacitor is governed by V = Q/C where Q is the charge and C is the capacitance of the capacitor. The capacitor has not been charged. As a result, the charge on the capacitor is approximately equal to zero coulombs and by the definition of the voltage equation given above V = 0/C = 0. Hence, the voltage across the terminals of the capacitor is zero volts. The rest can be understood by applying Ohms Law (V = I *R). The internal resistance of a CAPACITOR can be modeled as a short when initially connected to a power source. I hope this assists you understand your question.

Great explination

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