parallel voltage??

Question

calculate the resistance of the voltmeter?

diagram below
http://img265.imageshack.us/my.php?image=32936731hr6.jpg

Answer 1

You can replace the voltmeter resistance in parallel with the 15k resistance with their equivalent ( 1 / R equivalent = 1 / 15 k + 1 / R voltmeter ) and if the ampermeter doesn't have a resistance you can use Ohm's law.

Answer 2

Doing this solution in a simple way, is ohms law on the indicated voltage source and the reading on the ammeter, and you find a equivalent load in the series circuit of 20,000 or 20k ohms.

This fully shows that the parallel circuit of the voltmeter and 15 k resistor has an equivalent of 10k ohms.

Again simply using the formula for a parallel resistance - knowing one value of 15k and a unknown resistance of the voltmeter combine to 10k in parallel.

Finding that the voltmeter has a internal resistance of 30k ohm and connected across the 15k ohm gives you the 10k needed to have a 250 microamp flow.

It is absolutely believeable ( proven here ) that the indicated amperage would be realistic and a very simple number as a solution.
This is all basic high school freshman science.

There is no such thing as a voltmeter having infinite resistance. The common way of stating internal resistance in any given voltmeter is given in "ohms per volt". Better meters have higher ohms per volt and therefore less "loading" which alters the expected voltage values less upon connecting the meter to the circuit. A circuit of much higher impedance or resistance, in this example problem, would be seriously affected by this particular voltmeter used in the drawing. With these resistances, the shunt within the ammeter will be a extremely minor addition to overall circuit load resistance.

It is recommended to have a solid foundation of basics before calling ones self an engineer...walk before you fly : )

Answer 3

First of all, a voltmeter is considered to have infinite resistance. The circuit shown is a series circuit with the voltmeter testing the voltage drop across the first resistance of 15Kohms. The figure has the wrong values for starters. If source voltage is 5V and the sum of both resistances is 25,000 ohms, the total amount of current has to be 0.2 mA not 0.25 mA. Using the correct value of 0.2 mA, the voltmeter will read 3V. The other resistance will chew up the other 2V which leaves 0V returning to the battery as it should in a series circuit. Make sense?

Answer 4

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