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

Question: Calculate the potential drop across one of the 100 kΩ resistors in the given circuit and analyze the percentage deviation when a voltmeter with an internal resistance of 10 MΩ is connected in parallel with one of the resistors. Answer: The potential drop across one of the 100 kΩ resistors is 6.0 V. When a voltmeter with an internal resistance of 10 MΩ is connected in parallel, the percentage deviation is approximately 0.33%.

Step-by-step solution

Calculate the equivalent resistance of two resistors in series

Since the two 100 kΩ resistors are in series, we can find the total resistance (\(R_{total}\)) by simply adding their resistances: \(R_{total} = R_1 + R_2 = 100 \mathrm{k}\Omega + 100 \mathrm{k}\Omega = 200 \mathrm{k}\Omega\)

Calculate the current in the circuit

Now that we know the total resistance, we can use Ohm's Law to calculate the current (\(I\)) flowing in the circuit: \(I = \frac{V}{R_{total}} = \frac{12.0 \mathrm{V}}{200 \mathrm{k}\Omega} = 60 \mathrm{\mu A}\)

Calculate the potential drop across one resistor (part a)

Since the current is the same through both resistors, we can use Ohm's Law again to find the potential drop (\(V_{drop}\)) across one of the resistors: \(V_{drop}= I \times R_1 = 60\mathrm{\mu A} \times 100 \mathrm{k}\Omega= 6.0 \mathrm{V}\)

Calculate the equivalent resistance when a voltmeter is connected

When the voltmeter with an internal resistance of 10 MΩ is connected in parallel with one of the resistors, the equivalent resistance of this parallel combination (\(R_{equiv}\)) can be computed using the formula for resistors in parallel: \(\frac{1}{R_{equiv}} = \frac{1}{R_1} + \frac{1}{R_{voltmeter}}\) \(R_{equiv} = \frac{R_1 \times R_{voltmeter}}{R_1 + R_{voltmeter}} = \frac{100\mathrm{k}\Omega \times 10\mathrm{M}\Omega}{100\mathrm{k}\Omega + 10\mathrm{M}\Omega} \approx 99.01\mathrm{k}\Omega\)

Calculate the new total resistance and current

We now have one resistor with resistance \(R_{equiv}\) of about 99.01 kΩ and the other resistor of 100 kΩ still in series: \(R_{total,new} = R_{equiv} + R_2 \approx 99.01 \mathrm{k}\Omega + 100 \mathrm{k}\Omega \approx 199.01 \mathrm{k}\Omega\) The new current flowing in the circuit can be calculated using Ohm's Law: \(I_{new} = \frac{V}{R_{total,new}} \approx \frac{12.0 \mathrm{V}}{199.01 \mathrm{k}\Omega} \approx 60.24 \mathrm{\mu A}\)

Calculate the new potential drop across one resistor

Now the new potential drop (\(V_{drop,new}\)) across the resistor with the voltmeter connected in parallel can be found using Ohm's Law: \(V_{drop,new} = I_{new} \times R_{equiv} \approx 60.24\mathrm{\mu A} \times 99.01 \mathrm{k}\Omega \approx 5.98 \mathrm{V}\)

Calculate the percentage deviation (part b)

Finally, we will calculate the percentage deviation (∆%) between the potential drop without the voltmeter (\(V_{drop}\)) and the potential drop with the voltmeter (\(V_{drop,new}\)): \(\Delta \% = \frac{V_{drop} - V_{drop,new}}{V_{drop}} \times 100 = \frac{6.0 \mathrm{V} - 5.98 \mathrm{V}}{6.0 \mathrm{V}} \times 100 \approx 0.33 \%\) The voltmeter reading will deviate by approximately 0.33% from the value determined in part (a).