As shown in the figure, a 6.0000 -V battery is used to...

As shown in the figure, a 6.0000 -V battery is used to produce a current through two identical resistors, $R$, each having a resistance of $100.00 \mathrm{k} \Omega$. A digital multimeter (DMM) is used to measure the potential difference across the first resistor. DMMs typically have an internal resistance of $10.00 \mathrm{M} \Omega$. Determine the potential differences $V_{a b}$ (the potential difference between points $a$ and $b$, which is the difference the DMM measures) and $V_{b c}$ (the potential difference between points $b$ and $c$, which is the difference across the second resistor). Nominally, $V_{a b}=V_{b c}$, but this may not be the case here. How can this measurement error be reduced?

Key Concepts

Voltage Divider

A voltage divider is a fundamental circuit configuration where a voltage source is connected to a series of resistors, and the voltage is divided among the resistors proportionally to their resistances. This concept is essential as it forms the basis for understanding how voltage drops occur in series circuits and how different measurements across segments of the divider may be affected by changes in resistance values.

Internal Resistance of Measurement Instruments

Measuring devices, such as digital multimeters (DMMs), have an inherent internal resistance that ideally should be infinitely high to avoid drawing current from the circuit under test. In practice, the finite internal resistance loads the circuit and alters the effective resistance across which the voltage is measured, thus potentially introducing measurement error.

Loading Effect

The loading effect refers to the impact that a measurement instrument has on the circuit being measured due to its finite internal resistance. When a DMM is connected across a component in a circuit, it forms a parallel combination with that component, thereby changing the total resistance seen by the circuit. This alteration can lead to discrepancies in the voltage drop across components, especially in circuits where the component resistances are comparable to the DMM’s internal resistance.

Measurement Error Reduction Techniques

To reduce measurement error caused by the loading effect, one common approach is to use measuring instruments with much higher internal resistance compared to the circuit resistances. Alternatively, buffering the measurement point with a voltage follower or isolation amplifier can isolate the circuit from the load imposed by the measuring device, thereby preserving the original circuit conditions and yielding more accurate voltage readings.

Transcript

00:01 Hi, in the given problem there are two resistors identical resistors joined in series combination.

00:15 Each having a resistance are having a value 100 .00 kilo -oom.

00:25 Then a battery providing a potential of v.

00:32 Is equal to 6 .00 volt has been put across these two resistors and then across one of the resistors a multimeter is put which is having its internal resistance are r r i as 10 .00 mega -oam now first of all we have to find the potential drop taking place across these two resistors means across a -b and across b c.

01:18 Here this is c.

01:20 To find this potential drop first of all we should measure the total current passing through this circuit and to find the total current passing through the circuit we should have the total resistance, the value of the total resistance equivalent resistance of this circuit for which first of all this r i is in parallel with i is in parallel with r so as r i internal resistance of the multimeter means the multimeter is in parallel with this resistance so we can say as r i is in parallel with r so the net resistance let it be r p where p represents the parallel combination or not representing by p we can represent it by ab as this is the resistance across ab.

02:23 So this rab will be given by when two resistors are joined in parallel, we know their net resistance is given as their product in numerator and their addition in denominator.

02:37 So putting these values here.

02:39 First of all, r is 100 .00 kilo -oom.

02:45 Is 10 mega -oom or we can say 10 into 1 ,000 kilo -oom divided by again 100 .00 plus it will become 10 ,000 kilo -oam.

03:06 Cancelling this one kilo -oom, we get this net resistance across ab to be equal to 99 .0099 kilo om.

03:18 Now this r a b is in series with this r so equivalent resistance of this circuit now will be given by r a b plus r b c and here rab is 99 .00 .m plus r b c is 100 .00 kilo -oom...

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