A typical voltmeter has an internal resistance of 10 MO and can only measure voltage differences

A typical voltmeter has an internal resistance of 10 MO and...

A typical voltmeter has an internal resistance of 10 $\mathrm{MO}$ and can only measure voltage differences of up to several hundred volts. Figure 81 shows the design of a probe to measure a very large voltage difference $V$ using a voltmeter. If you want the voltmeter to read 50 $\mathrm{V}$ when $V=50 \mathrm{kV}$ what value $R$ should be used in this probe? $$ \begin{array}{l}{\text { FIGURE 81 }} \\ {\text { Problem } 92 .}\end{array} $$

A typical voltmeter has an internal resistance of 10 $\mathrm{MO}$ and can only measure voltage differences of up to several hundred volts. Figure 81 shows the design of a probe to measure a very large voltage difference $V$ using a voltmeter. If you want the voltmeter to read 50 $\mathrm{V}$ when $V=50 \mathrm{kV}$ what value $R$ should be used in this probe?
$$
\begin{array}{l}{\text { FIGURE 81 }} \\ {\text { Problem } 92 .}\end{array}
$$

Key Concepts

Instrumentation Input Resistance

Instrumentation input resistance, such as that of a voltmeter, plays an important role in measurement circuits. The finite internal resistance of measurement devices must be accounted for when designing circuits like voltage dividers, as it affects the overall resistance seen by the circuit and, consequently, the accuracy of voltage scaling.

Voltage Divider

A voltage divider is a circuit configuration that uses two or more resistors in series to scale down a higher voltage to a lower value in proportion to the resistor values. This principle is fundamental when adapting signals so that a measurement device, which can only handle lower voltages, is not exposed to excessive voltage levels.

High Voltage Measurement Techniques

High voltage measurement techniques involve converting high voltage signals to lower, measurable levels by using passive components such as resistor networks. This process allows standard instruments, which are typically rated for lower voltages, to indirectly measure high voltage differences safely and accurately.

Transcript

00:01 Let's say the internal resistance of the voltmeter be indicated by rv, and this is given to be 10 oms, and let the 15 mega -oom resistance be indicated by r15.

00:21 So we calculate the current through the probe and the voltmeter as the voltage across the probe.

00:29 So basically i, so we calculate the current through the probe and voltmeter.

00:35 So this will be equal to voltage across the probe divided by the equivalent resistance.

00:43 So this is basically from oomslop.

00:46 Now we need to know what the equivalent resistance will be in order to plug it in the denominator.

00:53 So we have r in series with the parallel combination of rv and r15.

01:04 So r is in series.

01:09 So we add this with the equivalent resistance of these two registers.

01:13 Now these two registers are in parallel.

01:16 So the equivalent resistance of the parallel combination is this.

01:26 Now vv, which is voltage across the internal resistors, is i times are the equivalent resistance of these two.

01:47 This is because rv and r15 are in parallel.

01:56 So the voltage across these two should be same.

02:02 So vv should be equal to v15.

02:07 And v15 by oombs law is current flowing through the resistor time the resistance and here we have to use the equivalent resistance and equivalent not this but equivalent resistance of the internal resistance and 15 mega own resistance so basically we are let me just show you it's better to draw a diagram so this is rv and this is are 15 and these two are in parallel and they are then in series with r.

03:09 So the current flowing is i which we already calculated and v is the potential difference from across these two points.

03:26 So this is v and vv is the potential difference of this resistor but since these two are in parallel so the voltage vv will be equal to v15 as i wrote down as i wrote over here now you can write this to be equal to so you can write vv to be equal to v15 and by oom's law you can write this to be equal to ivr or i15 r15 but you don't know what iv is and you don't know what i -15 is but you do know what the total current is that's flowing through the circuit so instead of finding going with this method what you can do you can choose to take the total current and replace these two register by the equivalent resistor which has a magnitude of this and multiply it with the current.

04:42 So that will give you the same answer...

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