3. Suppose a multi-meter is used to measure voltage differences in the circuit b

Question

3. Suppose a multi-meter is used to measure voltage differences in the circuit below, represented as a schematic diagram. Suppose the battery is 10 volts and each resistor is 47 mega Ohms. (a) without applying the meter what should be the voltage difference between points b and c? (b) If you want to measure the voltage difference from point b to point a (a positive value) you connect the vo wire and where would you connect the COM wire? Indicate these connections with lines. v OFF to a O O O O Com to b VO COM A (c) Now suppose you actually make this measurement and find V, -V, 4.89 Volts What is the resistance of the volt meter? (Actually, the effective volt-meter resistance will depend on which scale you have it set to, so this would be for a particular meter setting.) 6-3 UND Physics CPSL

Explanation / Answer

3a) Since all the resistances are in series. so equivalent resistance will just add up.

Eq R = 47+47 = 94 mega om

Current i = V/R = 10/94000000 = 1.06X10^-7

So, voltage diff between b and c Del V = IXR = 1.06X10^-7X47X10^6 =4.982 V

3b) At points b and c respectively.

3c) If resistance of voltmetre is Rv then R2 and rv will be in parralel.

As voltage drop in R2 is given as 4.89,

so current through R2 i2 = 4.89/43X10^-6 = 1.1X10^-7

But total resistance now Rt = 43 + 43Rv/(43+rv)

So, current from here i2 = V/Rt = 10/Rt = 1.1X10^-7 from this equation Rv can be easily calculated.

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