High voltage circuit with voltmeter in between

[ddobre]

Homework Statement

You have a voltmeter with an internal resistance of 10 MΩ. You would like to measure a very large voltage source, but you notice that this overwhelms your voltmeter, and you begin to get inaccurate results if the voltage is too high. You design the circuit shown above as a workaround. When measuring a voltage source of 15 kV, you would like your voltmeter to read 50 V. What value of R must you use to achieve this?

Homework Equations

Vab = (emf) - Ir

V = IR

(Parallel): 1/Req = 1/R1 + 1/R2

(Series): Req = R1 + R2

The Attempt at a Solution

The first thing I tried was to calculate current with Vab = 50v, emf = 15000v, and r = 10MΩ, to which I got I = .001495A . After calculating this I got a little confused as to which equation to use next because of the location of the voltmeter. I know that in series the current remains the same, but also the voltmeter is kind of like a resistor too, one which is in parallel with the 15MΩ resistor. One attempt of mine was to calculate Req for the parallel situation of the voltmeter and the 15MΩ resistor, with Req = 6000000Ω. From there, I tried setting up a series equation with the Req just calculated as a single resistor in series with the R resistor I am looking for. But this is where I got confused. I tried equating different values using V1/R1 = V2/R2: (50v)/(6000000Ω) = (15000v)/R, getting R = 1800000000Ω. But, I think this is out of proportion with the rest of the circuit. Any advice?

 

[Baluncore]

You are close but have forgotten the last step subtraction.

Use SI multipliers not strings of zeros.
Replace the decimal point with the multiplier.
10M0 meter in parallel with 15M0 = 6M0
Current needed is then 50V / 6M0 = 8u333 A
15k V / 8u333 A = 1G8 ohms total.
But must subtract the 6M0 so R = 1G8 – 6M0 = 1G794 ohms = 1794 Megohm.

 

[ddobre]

You are close but have forgotten the last step subtraction.

Use SI multipliers not strings of zeros.
Replace the decimal point with the multiplier.
10M0 meter in parallel with 15M0 = 6M0
Current needed is then 50V / 6M0 = 8u333 A
15k V / 8u333 A = 1G8 ohms total.
But must subtract the 6M0 so R = 1G8 – 6M0 = 1G794 ohms = 1794 Megohm.

Ah, so some of my work is okay. But why do I subtract the Req = 6M0 from the 1G8? I'm assuming it has something to do with the parallel situation, because resistance goes down with resistors in parallel? Or does that not really apply here to this subtraction?

 

[Baluncore]

The common reference voltage is the bottom line.
Resistors in series add. The voltages across series resistors add.

The meter and parallel resistor are equivalent to a 6M0 resistor. Call it Rp.
The high value series resistor is Rs.
Together, Rs and Rp make a potential divider.

Since the same current flows through both Rs and Rp we have;
current = ( Rs + Rp ) / 15 kV = Rp / 50 V
Rs + Rp = Rp * 15 kV / 50 V
Rs = ( Rp * 15 kV / 50 V ) – Rp