A circuit consists of a series combination of $\mathbf{6}\mathbf{K\Omega }$and $\mathbf{5}\mathbf{K\Omega }$resistors connected across a 50 vbattery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the$\mathbf{5}\mathbf{K\Omega }$resistor using a voltmeter having an internal resistance of ${\mathbf{R}}_{\mathbf{c}}\mathbf{=}\mathbf{10}\mathbf{K\Omega }$. (a) What potential difference does the voltmeter measure across the$\mathbf{5}\mathbf{K\Omega }$resistor? (b) What is the truepotential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in errorrom the true potential difference?

Potential difference between any two points is defined as the amount of work done in moving a unit charge from one point to another

Given
We are given two resistors in series$6\mathrm{K\Omega }5\mathrm{K\Omega }$acrossabatteryAnd we have a voltmeter with internalresistance, this internal resistance is the resistance of the coil${\mathrm{R}}_{\mathrm{c}}=10\mathrm{K\Omega }$
Solution
(a) The voltmeter is connected across$5\mathrm{K\Omega }$so it measures the voltage betWeen the terminals of this resistance. But in this
case, the internal resistance of the voltmeter is in parallel with the resistance$5\mathrm{K\Omega }$ as shown in the figure below. The
current flows in this combination of the resistors, where the equivalent resistance of two resistors in parallel is given by
equation 26.3 in the next form
${\mathrm{R}}_{\mathrm{c}}=10\mathrm{K\Omega }$

$$\begin{gathered} {\mathrm{R}}_{\mathrm{e}}=\frac{5\left({\mathrm{R}}_{\mathrm{c}}\right)}{5+{\mathrm{R}}_{\mathrm{c}}} \\ {\mathrm{R}}_{\mathrm{eq}}=\frac{5\times 10}{5+10} \\ =3.33\mathrm{k\Omega } \end{gathered}$$

The voltmeter will read the voltage due to this combination of resistance and it is given by Ohm's law in the form

$\mathrm{V}={\mathrm{IR}}_{\mathrm{eq}}$

Where I is the current flow due to the battery. The current flow in 6kiloohms is the same for the equivalent resistance 3.33 kohms as both resistors are in series and We can measure this current using Ohm's law in the next form

$$\begin{gathered} \mathrm{l}=\frac{\hat{\mathrm{i}}}{6+3} \\ =\frac{50\mathrm{V}}{6\mathrm{k\Omega }+3.33\mathrm{K\Omega }} \\ =0.00536\mathrm{A} \end{gathered}$$

NowWecanplugourvaluesforIandReqintoequation(1)togetV
$$\begin{gathered} \mathrm{V}={\mathrm{IR}}_{\mathrm{eq}} \\ =0.00536\mathrm{A}\times 3.33\mathrm{K\Omega } \\ =17.85\mathrm{V} \end{gathered}$$
Therefore the voltage across 5.00 kohms and measured by the voltmeter.17.85V

(b)Thetruepotentialisthevoltageduetoonlytheresistance5.00kohmsWhereRcisabsence-SoWecanuseOhm'slaw
again to measure this potential
$\mathrm{V}=\mathrm{I}\left(5.00\mathrm{k\Omega }\right)$
Where I is the current floWs due to the battery. This current floWs in 6.00 kohms is the same for the resistance 5.00 kohms as bo
resistors are in series and We can measure this current using Ohm's law in the next form
resnsmrs are In series ano we can measure InlS current usung unm‘s Law In me nex: worm
$$\begin{gathered} \mathrm{I}=\frac{\hat{\mathrm{I}}}{6\mathrm{K\Omega }+5\mathrm{\Omega }} \\ =\frac{50}{6+5} \\ =0.00450\mathrm{A} \end{gathered}$$

NowwecanplugourvaluesforIandReqintoequation(1)togetV
V = I (5)

$$\begin{gathered} =0.00450\mathrm{A}\left(5\mathrm{k\Omega }\right) \\ =22.72\mathrm{V} \end{gathered}$$

Therefore the Potential difference across the resistor is 22.72V

(c)Thereisanerrorinthereadingofpotentialduetotheresistanceofmeter.Thedifferencebetweenthevoltagemeasured
by the meter and the true voltage is

$$\begin{gathered} ∆={\mathrm{V}}_{\mathrm{true}}-{\mathrm{V}}_{\mathrm{meter}} \\ =22.72\mathrm{V}-17.85\mathrm{V} \\ =4.87\mathrm{V} \end{gathered}$$
The error represents, the ratio between this difference and the true potential
$$\begin{gathered} \%\mathrm{error}=\frac{∆}{22.72} \\ =21.43\% \end{gathered}$$

Therefore the percentage error is 21.43%