Question

A circuit consists of two \(100 .-\mathrm{k} \Omega\) resistors in series with an ideal \(12.0-\mathrm{V}\) battery. a) Calculate the potential drop across one of the resistors. b) A voltmeter with internal resistance \(10.0 \mathrm{M} \Omega\) is connected in parallel with one of the two resistors in order to measure the potential drop across the resistor. By what percentage will the voltmeter reading deviate from the value you determined in part (a)? (Hint: The difference is rather small so it is helpful to solve algebraically first to avoid a rounding error.)

Short answer

Answer: The potential drop across one resistor in the series circuit is 6.0 V. When a voltmeter is connected in parallel to one of the resistors, the voltmeter reading deviates from the value determined by approximately 0.33%.

Step-by-step solution

Find the total resistance of the series circuit

First, we will find the total resistance of the two \(100\,\text{k}\Omega\) resistors in series. The formula for resistors in series is: \(R_{total} = R_1 + R_2\) where \(R_{total}\) is the total resistance, and \(R_1\) and \(R_2\) are the resistances of the individual resistors. \(R_{total} = 100\,\text{k}\Omega + 100\,\text{k}\Omega = 200\,\text{k}\Omega\) The total resistance of the series circuit is \(200\,\text{k}\Omega\).

Calculate the current in the circuit

Now that we have the total resistance, we can use Ohm's law to find the current in the circuit. Ohm's law states that: \(V = IR\) where \(V\) is the potential difference across the resistors, \(I\) is the current, and \(R\) is the resistance. We can rearrange the equation to solve for the current: \(I = \frac{V}{R}\) \(I = \frac{12.0\,\text{V}}{200\,\text{k}\Omega} = 60\,\mu\text{A}\) The current in the circuit is \(60\,\mu\text{A}\).

Calculate the potential drop across one resistor

Since the current is the same in a series circuit, we can use Ohm's law again to find the potential drop across one resistor: \(V_{drop} = IR\) \(V_{drop} = (60\,\mu\text{A})(100\,\text{k}\Omega) = 6.0\,\text{V}\) The potential drop across one resistor is \(6.0\,\text{V}\). This is the answer to part (a).

Find the equivalent resistance of the parallel connection

Now we will find the equivalent resistance of the parallel connection of one resistor and the voltmeter. The formula for two resistors in parallel is: \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\) We can plug in the values of the resistances: \(\frac{1}{R_{eq}} = \frac{1}{100\,\text{k}\Omega} + \frac{1}{10\,\text{M}\Omega}\) To find \(R_{eq}\), we need to solve for it: \(R_{eq}=\frac{1}{\left(\frac{1}{100\,\text{k}\Omega}+\frac{1}{10\,\text{M}\Omega}\right)} \approx 99.01\,\text{k}\Omega\) The equivalent resistance of the parallel connection is approximately \(99.01\,\text{k}\Omega\).

Calculate the new potential drop across the parallel connection

Now that we have the equivalent resistance, we can use Ohm's law to find the new potential drop across the parallel connection. First, we need to find the total resistance of the series circuit with the parallel connection: \(R'_{total}=100\,\text{k}\Omega + 99.01\,\text{k}\Omega \approx 199.01\,\text{k}\Omega\) Now, we can find the new current in the circuit using Ohm's law: \(I' = \frac{12.0\,\text{V}}{199.01\,\text{k}\Omega} \approx 60.30\,\mu\text{A}\) Finally, we can find the new potential drop across the parallel connection: \(V'_{drop} = (60.30\,\mu\text{A})(99.01\,\text{k}\Omega) \approx 5.98\,\text{V}\) The new potential drop across the parallel connection is approximately \(5.98\,\text{V}\).

Calculate the percentage deviation of the voltmeter reading

Finally, we can calculate the percentage deviation of the voltmeter reading as follows: \(percentage\,deviation = \frac{V_{drop} - V'_{drop}}{V_{drop}} \times 100\%\) \(percentage\,deviation = \frac{6.0\,\text{V} - 5.98\,\text{V}}{6.0\,\text{V}} \times 100\% \approx 0.33\%\) The voltmeter reading deviates from the value determined in part (a) by approximately \(0.33\%\). This is the answer to part (b).