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Two capacitors in series are charged through a resistor. Identical capacitors are instead connected in parallel and charged through the same resistor. How do the times required to fully charge the two sets of capacitors compare?

Short Answer

Expert verified
Answer: Capacitors connected in series will fully charge faster compared to those connected in parallel. The time required to fully charge capacitors in series is one-fourth the time required for capacitors in parallel when charged through the same resistor.

Step by step solution

01

Determine the equivalent capacitance of capacitors in series and parallel configurations.

When capacitors are connected in series, their equivalent capacitance is given by the formula: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) When capacitors are connected in parallel, the equivalent capacitance is given by the formula: \(C_{eq} = C_1 + C_2\) Assuming both capacitors have the same capacitance \(C\) in each configuration, let's find the equivalent capacitance for both scenarios.
02

Calculate the equivalent capacitance for capacitors in series.

For capacitors in series: \(\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C}\) \(\frac{1}{C_{eq}} = \frac{2}{C}\) \(C_{eq(series)} = \frac{C}{2}\)
03

Calculate the equivalent capacitance for capacitors in parallel.

For capacitors in parallel: \(C_{eq(parallel)} = C + C\) \(C_{eq(parallel)} = 2C\)
04

Determine the time constant for each configuration.

The time constant for charging a capacitor through a resistor is given by the formula: \(\tau = R C_{eq}\) Here, \(R\) is the resistance through which capacitors are charged. The time required to fully charge the capacitors (to about 99%) is around 5 times the time constant, i.e. \(t_{charge} \approx 5\tau\). Let's calculate the time constant for each configuration.
05

Calculate the charging time for capacitors in series.

For capacitors in series: \(\tau_{series} = R C_{eq(series)} = R \cdot \frac{C}{2}\) \(t_{charge(series)} \approx 5 \tau_{series} = 5R \cdot \frac{C}{2}\)
06

Calculate the charging time for capacitors in parallel.

For capacitors in parallel: \(\tau_{parallel} = R C_{eq(parallel)} = R \cdot 2C\) \(t_{charge(parallel)} \approx 5 \tau_{parallel} = 5R \cdot 2C\)
07

Compare the times required to fully charge the capacitors.

From Steps 5 and 6, we found that: \(t_{charge(series)} = 5R \cdot \frac{C}{2}\) \(t_{charge(parallel)} = 5R \cdot 2C\) The ratio of charging time between the two configurations is: \(\frac{t_{charge(series)}}{t_{charge(parallel)}} = \frac{5R \cdot \frac{C}{2}}{5R \cdot 2C} = \frac{1}{4}\) Thus, the time required to fully charge capacitors in series is one-fourth the time required for capacitors in parallel when charged through the same resistor.

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Most popular questions from this chapter

How long would it take, in multiples of the time constant, \(\tau,\) for the capacitor in an \(\mathrm{RC}\) circuit to be \(98 \%\) charged? a) \(9 \tau\) c) \(90 \tau\) e) \(0.98 \tau\) b) \(0.9 \tau\) d) \(4 \tau\)

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.)

How long will it take for the current in a circuit to drop from its initial value to \(1.50 \mathrm{~mA}\) if the circuit contains two \(3.8-\mu \mathrm{F}\) capacitors that are initially uncharged, two \(2.2-\mathrm{k} \Omega\) resistors, and a \(12.0-\mathrm{V}\) battery all connected in series?

You want to make an ohmmeter to measure the resistance of unknown resistors. You have a battery with voltage \(\mathrm{V}_{\mathrm{emf}}=9.00 \mathrm{~V}\), a variable resistor, \(R,\) and an ammeter that measures current on a linear scale from 0 to \(10.0 \mathrm{~mA}\) a) What resistance should the variable resistor have so that the ammeter gives its full-scale (maximum) reading when the ohmmeter is shorted? b) Using the resistance from part (a), what is the unknown resistance if the ammeter reads \(\frac{1}{4}\) of its full scale?

The dead battery of your car provides a potential difference of \(9.950 \mathrm{~V}\) and has an internal resistance of \(1.100 \Omega\). You charge it by connecting it with jumper cables to the live battery of another car. The live battery provides a potential difference of \(12.00 \mathrm{~V}\) and has an internal resistance of \(0.0100 \Omega\), and the starter resistance is \(0.0700 \Omega\). a) Draw the circuit diagram for the connected batteries. b) Determine the current in the live battery, in the dead battery, and in the starter immediately after you closed the circuit.

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