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How to calculate values from amp meter and volt meter?

  • http://i53.tinypic.com/20atea1.jpg?tti Ok I have the values for current and voltage for the individual resistors. Now I need to figure out how to calculate what the amp meter and volt meter will calculate in order to find out which configuration will be the most accurate (within 3 to 4 decimal places, how do I do this for the two circuits with the amp and voltmeter connected?. Total R = 200 I = .03 Voltage for the 20 ohm resistor is .6 Voltage for the 180 ohm resistor is 5.4

  • Answer:

    Just know that the resistance of the voltmeter is connected in parallel to it and that of the amp meter is in series with it. Just put 20000 ohm resistance parallel to the voltmeter and 0.5 ohm resistance in series with the amp meter. Now solve the circuits to find which one is accurate. To me the first connection is more accurate.

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When no meter is used, current is simply 6V/(180+20) = 30 mA. It is 30.00000000mA!! In the next circuit where the voltmeter is right across 180, the resistance of 180||20000 = 178.3944 ohms. This is in series with 20 and 0.5 of ammeter. That makes the total as 198.89445.Thus current is 30.1668mA. This is 0.1668mA more than expected actual current. The voltage across 180 ohms is 5.3816V. This is lesser than expected value of 5.4 by 18.4mV. Do similar calculations in next circuit and see what differences are there in that circuit from 5.4V and 30.000mA. You will find the current to be 29.90mA and voltage as 5.397V. Obviously this circuit is one that is more accurate. The current will be 6/{20+180.5||20000} = 30.17 in the 20 ohms. The current in 0.5 ohms will be however 29.90mA.

When no meter is used, current is simply 6V/(180+20) = 30 mA. It is 30.00000000mA!! In the next circuit where the voltmeter is right across 180, the resistance of 180||20000 = 178.3944 ohms. This is in series with 20 and 0.5 of ammeter. That makes the total as 198.89445.Thus current is 30.1668mA. This is 0.1668mA more than expected actual current. The voltage across 180 ohms is 5.3816V. This is lesser than expected value of 5.4 by 18.4mV. Do similar calculations in next circuit and see what differences are there in that circuit from 5.4V and 30.000mA. You will find the current to be 29.90mA and voltage as 5.397V. Obviously this circuit is one that is more accurate. The current will be 6/{20+180.5||20000} = 30.17 in the 20 ohms. The current in 0.5 ohms will be however 29.90mA.

veeyesvee

Just know that the resistance of the voltmeter is connected in parallel to it and that of the amp meter is in series with it. Just put 20000 ohm resistance parallel to the voltmeter and 0.5 ohm resistance in series with the amp meter. Now solve the circuits to find which one is accurate. To me the first connection is more accurate.

Imrul Ahsan

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