An ammeter must be placed in series with the branch of the circuit you are tryin

Question

An ammeter must be placed in series with the branch of the circuit you are trying to measure the current of, and consequently an ideal ammeter has an internal resistance equal to 0 Ohms. This means that an ideal ammeter will not alter the resistance of any circuit into which it is inserted. A voltmeter is placed in parallel with the device that you are trying to measure the voltage difference across. An ideal voltmeter will have an infinite resistance so as not to create another branch of the circuit for current to flow through. Real ammeters have some small resistance and therefore alter the total current flowing through the circuit. Real voltmeters have a less than infinite resistance and also affect the circuit being tested when they are connected. Because a real voltmeter has a resistance that is not infinite, in the circuit shown in figure 1 (in the manual), some of the current, IA measured by the ammeter, A. will pass through the voltmeter instead of through R. Let this current be Iv. Assume an internal resistance for the voltmeter of RV and assume that the voltmeter ascts like a pure resistor and will obey Ohm's Law. Show that IA = I(1 + R/RV) Show that for I to be within 1% of IA that RV must be greater than 100R. The resistance of the DVM is 107 Ohms-what is the percent difference between the value of R using I and IA?

Explanation / Answer

Voltmeter is parallel to resistor ; so Rv is parallel to R ; so finding Equivalent Resistance by 1/Req = 1/Rv + 1/R ; Req = Rv*R / (Rv + R) ; Ia get distributed into two components ; Iv = Vv /Rv ; Vv = Ia*Req ; and so Iv = Vv/Rv = Ia*R/(R + Rv) ; so Ia = Iv*(R+Rv)/R = Iv*(1 + Rv/R); ------ (Proved) Now for Iv = 0.01*Ia ; Ia = 0.01*Ia * (1 + Rv/R) 100 = 1 + Rv/R Rv/R = 99 Rv = 99 R ; -------(Proved) So Rv should be greater than 99*R or 100*R

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