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Chapter 26: Problem 21

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

An ammeter with an internal resistance of \(53 \Omega\) measures a current of \(5.25 \mathrm{~mA}\) in a circuit containing a battery and a total resistance of \(1130 \Omega\). The insertion of the ammeter alters the resistance of the circuit, and thus the measurement does not give the actual value of the current in the circuit without the ammeter. Determine the actual value of the current.

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) \(4 \tau\)

Kirchhoff's Junction Rule states that a) the algebraic sum of the currents at any junction in a circuit must be zero. b) the algebraic sum of the potential changes around any closed loop in a circuit must be zero. c) the current in a circuit with a resistor and a capacitor varies exponentially with time. d) the current at a junction is given by the product of the resistance and the capacitance. e) the time for the current development at a junction is given by the product of the resistance and the capacitance.

Which of the following has the same unit as the electromotive force (emf)? a) current b) electric potential c) electric field d) electric power e) none of the above

You wish to measure both the current through and the potential difference across some component of a circuit. It is not possible to do this simultaneously and accurately with ordinary voltmeters and ammeters. Explain why not.
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