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(II) Two capacitors connected in parallel produce an equivalent capacitance of 35.0 \(\mu\)F but when connected in series the equivalent capacitance is only 4.8 \(\mu\)F. What is the individual capacitance of each capacitor?

Short Answer

Expert verified
The capacitances are 5.0 µF and 30.0 µF.

Step by step solution

01

Understand the Equations for Capacitors in Parallel and Series

When two capacitors, say \( C_1 \) and \( C_2 \), are connected in parallel, the equivalent capacitance \( C_p \) is given by the formula:\[ C_p = C_1 + C_2 \]. In this problem, \( C_p = 35.0 \, \mu\text{F} \). When they are connected in series, the equivalent capacitance \( C_s \) is given by the formula:\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]. Here, \( C_s = 4.8 \, \mu\text{F} \).
02

Set Up the Equations

We have two equations from the above understanding:\[ C_1 + C_2 = 35.0 \].\[ \frac{1}{4.8} = \frac{1}{C_1} + \frac{1}{C_2} \]. These two equations are the starting point for calculating the individual capacitances.
03

Simplify the Series Equation

Rearrange the series equation: \[ \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{4.8} \]. Let \( C_1 + C_2 = 35 \), so assume \( C_1 = 35 - C_2 \), and substitute into the series equation: \[ \frac{1}{4.8} = \frac{1}{C_1} + \frac{1}{C_2} \].
04

Solve the System of Equations

Substitute \( C_1 = 35 - C_2 \) into the series equation which becomes:\[ \frac{1}{4.8} = \frac{1}{(35 - C_2)} + \frac{1}{C_2} \]. Solve for \( C_2 \) by rearranging terms and solving the resulting quadratic equation.
05

Solve the Quadratic Equation for Capacitors

Multiply the equation by \( (35 - C_2)C_2 \) to clear fractions and obtain a quadratic equation: \[ 35C_2 - C_2^2 = 4.8C_2 \]. Simplify to \( C_2^2 - 30.2C_2 + 35 \). Use the quadratic formula: \( C_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = -30.2, c = 35 \).
06

Find Values of C1 and C2

Calculate the discriminant: \( 30.2^2 - 4 \cdot 1 \cdot 35 = 912.04 - 140 \), which is positive, ensuring real solutions. Solving for \( C_2 \), we get two possible values: approximately \( C_2 = 30.0 \) or \( C_2 = 5.0 \). Substituting back, we find \( C_1 \) as approximately \( 5.0 \) or \( 30.0 \).
07

Confirm the Solutions

The calculated values satisfy \( C_1 + C_2 = 35 \) and the series equation, as verified by checking \( \frac{1}{4.8} = \frac{1}{5} + \frac{1}{30} \). Thus, the solutions are \( C_1 = 5.0 \text{ and } C_2 = 30.0 \) \(\mu\text{F}\) (or vice versa).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalent Capacitance
When working with capacitors, it's essential to understand the concept of equivalent capacitance. Equivalent capacitance refers to the total capacitance of a network of capacitors connected together in either series or parallel. This value helps simplify complex circuits by allowing these groupings of capacitors to be treated as a single capacitor.
For capacitors in series, the formula used to find the equivalent capacitance, denoted as \( C_s \), is:
\[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \, ... \,.\]This indicates that the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. This leads to a smaller equivalent capacitance than any single capacitor in the series.
In contrast, for capacitors connected in parallel, the equivalent capacitance, \( C_p \), is straightforwardly the sum of each capacitor's capacitance:
\[C_p = C_1 + C_2 + C_3 + ... \,.\]In this setup, the equivalent capacitance is always greater than the largest single capacitance in the circuit. Both these concepts are pivotal when analyzing circuits with multiple capacitors as they determine how electric charge storage capacity is influenced by different configurations.
Capacitors in Parallel
Connecting capacitors in parallel is a common technique to increase the total capacitance in a circuit. When capacitors are connected in parallel, their total or equivalent capacitance is the simple sum of the individual capacitances:
\[C_p = C_1 + C_2 + C_3 + \, ... \,.\]In practical terms, each capacitor provides an additional path for charge to flow, effectively combining their storage capabilities. This directly increases the circuit’s ability to store energy.

Some helpful points to remember:
  • The voltage across each capacitor in parallel is the same as the voltage across the entire group because they are connected to the same two nodes in the circuit.
  • This arrangement results in a combined capacitance greater than any individual capacitor, making it useful for designing circuits that require large capacitance.
  • You might use this arrangement to ensure there is enough energy storage for devices that experience short bursts of high-energy demands.
In summary, capacitors in parallel are helpful when you need to increase a circuit's capacitance without increasing the voltage rating.
Capacitors in Series
When capacitors are connected in series, their total or equivalent capacitance is found using the formula:
\[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \, ... \,.\]This equation means that the reciprocal of the total capacitance is equal to the sum of the reciprocals of each capacitor's capacitance in the series.

Some key notes about capacitors in series:
  • The total capacitance of capacitors in series is typically less than the smallest individual capacitor in the network.
  • The charge stored is the same on each capacitor, but the voltage across each may vary depending on its capacitance. The sum of these voltages equals the total applied voltage.
  • This configuration is often used to increase the voltage rating of a circuit without significantly changing capacitance, as it spreads the voltage across the capacitors.
  • These can be ideal in filtering applications and can aid in distribution of electrical stress to prevent individual component failure.
Overall, connecting capacitors in series is beneficial for high-voltage applications where you seek greater insulation without requiring a single capacitor of high voltage rating.

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Most popular questions from this chapter

(I) Calculate the terminal voltage for a battery with an internal resistance of 0.900 \(\Omega\) and an emf of 6.00 V when the battery is connected in series with (\(a\)) a 71.0-\(\Omega\) resistor, and (\(b\)) a 710-\(\Omega\) resistor.

(II) An ammeter whose internal resistance is 53 \(\Omega\) reads 5.25 mA when connected in a circuit containing a battery and two resistors in series whose values are 720 \(\Omega\) and 480 \(\Omega\). What is the actual current when the ammeter is absent?

(II) A circuit contains a single 250-pF capacitor hooked across a battery. It is desired to store four times as much energy in a combination of two capacitors by adding a single capacitor to this one. How would you hook it up, and what would its value be?

A \(\textbf{Wheatstone bridge}\) is a type of "bridge circuit" used to make measurements of resistance. The unknown resistance to be measured, \(R_x,\) is placed in the circuit with accurately known resistances \(R_1, R_2,\) and \(R_3\) (Fig. 19-76). One of these, is a variable resistor which is adjusted so that when the switch is closed momentarily, the ammeter A shows zero current flow. The bridge is then said to be balanced. (\(a\)) Determine \(R_x\) in terms of \(R_1, R_2,\) and \(R_3\). (\(b\)) If a Wheatstone bridge is "balanced" when \(R_1 =\) 590 \(\Omega\), \(R_2 =\) 972 \(\Omega\), and \(R_3 =\) 78.6 \(\Omega\), what is the value of the unknown resistance?

(I) (\(a\)) Six 4.8-\(\mu\)F capacitors are connected in parallel. What is the equivalent capacitance? (\(b\)) What is their equivalent capacitance if connected in series?

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