A voltmeter is also a device that shows the voltage difference across its termin

Question

A voltmeter is also a device that shows the voltage difference across its terminals, but it has a very large resistance. The next two problems show why.

Suppose instead, the voltmeter were in parallel with the resistor as shown in the figure above.

Suppose a voltmeter having a resistance of 1500? were in series with a resistor of 50.0? as shown in the figure above.

1. If the system is connected up to a 12V battery, what would the voltage drop be across the voltmeter?
2. What would the voltage drop be across the 50? resistor? Is this voltmeter giving a good measure of the voltage drop across the 50? resistor? Explain why or why not.
3. Is this voltmeter giving a good measurement of what the voltage drop across the resistor would be if the voltmeter weren't there?

Suppose instead, the voltmeter were in parallel with the resistor as shown in the figure above.

1. If the system is connected up to a 12V battery, what would the voltage drop be across the meter? (Assume you can ignore the internal resistance of the battery.)
2. What is the voltage drop across the 50? resistor? Is this voltmeter giving a good measure of the voltage drop across the 50? resistor? Explain why or why not.
3. Is this voltmeter giving a good measurement of what the voltage would be across the resistor if the voltmeter weren't there?

Explanation / Answer

1. The equivalent resistance of the voltmeter and the resistor will be 1500+50=1500 ohms now current through the circuit will be =12/1550=.00774A drop across voltmeter=IR=(.00774)(1500)=11.612V 2. voltage drop across the resistor= IR= (.00774)*(50)=.387V The voltmeter is showing the drop across itself and not across the resistor, so it is not accurately displaying the voltage across the resistor. 3. if the voltmeter wont have been there then all 12V would have dropped across the resistor, but the voltmeter is showing 11.6V which is not accurate. case2: 1.the equivalent resistance will be= 50*1500/(50+1500)=48.38 ohms the current through the circuit will be = V/R= 12/48.38=.2480A and current through the voltmeter will be= I*50/(50+1500)=.008A and current through the resistor will be= .248-.008=.24A Drop across the voltmeter will be= IR=12V 2. drop across the resistor will be =IR= .24*50=12V the voltmeter gave a true measurement across the resistor, that is why because voltmeter is measuring the voltage across the resistor, and equivalent resistance is almost equal to 50 ohms(across which we want to measure the voltage) 3. had voltmeter not been there then the drop across the resistor will be 12V, and the voltmeter is showing the accurate result.

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