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Chapter 19: Problem 96

A 175-pF capacitor is connected in series with an unknown capacitor, and as a series combination they are connected to a 25.0-V battery. If the 175-pF capacitor stores 125 pC of charge on its plates, what is the unknown capacitance?

Short Answer

Expert verified
The unknown capacitance is approximately 5.15 pF.

Step by step solution

01

Understand the charge stored

In a series circuit, the charge stored by capacitors is the same. So, if the 175-pF capacitor stores 125 pC, this is the same charge stored by the unknown capacitor as well.
02

Use the formula for charge and capacitance

The relationship between charge \( Q \), capacitance \( C \), and voltage \( V \) for a capacitor is given by the formula \( Q = C \times V \). Rearrange this formula to solve for the voltage across the 175-pF capacitor: \( V = \frac{Q}{C} \).
03

Calculate the voltage across the 175-pF capacitor

Given \( Q = 125 \) pC and \( C = 175 \) pF, plug these values into the formula to find the voltage: \( V = \frac{125 \times 10^{-12}}{175 \times 10^{-12}} \approx 0.714 \) V.
04

Determine the voltage across the unknown capacitor

The total voltage supplied by the battery is 25.0 V. Subtract the voltage across the 175-pF capacitor to find the voltage across the unknown capacitor: \( 25.0 - 0.714 = 24.286 \) V.
05

Use the formula to find the unknown capacitance

Using the same formula, \( Q = C \times V \), rearrange to find the unknown capacitance: \( C = \frac{Q}{V} \). Substitute \( Q = 125 \text{ pC} \) and \( V = 24.286 \text{ V} \) to get \( C = \frac{125 \times 10^{-12}}{24.286} \approx 5.15 \times 10^{-12} \text{ F} \) or 5.15 pF.

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Key Concepts

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

Capacitance Calculation
When capacitors are connected in series, it alters the way we calculate their total capacitance. Unlike resistors in series where we simply add resistances, capacitors in series require a different approach. The formula to find the effective capacitance (\( C_{ ext{total}} \)) for capacitors in series is:\[\frac{1}{C_{ ext{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n}\]This means that the overall capacitance is actually less than any of the individual capacitances in the series.
To solve for an unknown capacitor in a series, we separately determine the voltage across known capacitors and use this to find the voltage across the unknown one.
Charge Storage
In a series connection, one interesting feature is that the charge (\( Q \)) stored in each capacitor is the same. This occurs because the same current flows through all the capacitors in the series.
Thus, if a given capacitor in the series stores a specific amount of charge, we can be confident that every other capacitor in the series stores this identical charge amount.
  • This is a crucial concept because it simplifies calculations. Instead of recomputing charge for each capacitor, you know it!
  • This consistency in charge regardless of individual capacitances is what you must remember while working with capacitors in series.
Voltage Distribution
Voltage distribution across capacitors in series can initially seem complex, but it's just about logical steps. The total supplied voltage from the battery is shared among the capacitors in the series.
First, you determine the voltage across a known capacitor using the formula:\[V = \frac{Q}{C}\]From here, subtract this voltage from the total voltage supplied by the battery to find the remaining voltage, which is across the unknown capacitor.
For example, with a total voltage of 25 V and finding a known voltage of 0.714 V, the remaining 24.286 V is across the unknown capacitor.
This calculated voltage is then the key to discovering the unknown capacitance.

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Most popular questions from this chapter

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?

(II) A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel. What is the battery's internal resistance?

(II) Three 1.70-k\(\Omega\) resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What are these four ways, and what is the net resistance in each case?

How much energy must a 24-V battery expend to charge a 0.45-\(\mu\)F and a 0.20-\(\mu\)F capacitor fully when they are placed (\(a\)) in parallel, (\(b\)) in series? (\(c\)) How much charge flowed from the battery in each case?

(II) The capacitance of a portion of a circuit is to be reduced from 2900 pF to 1200 pF.What capacitance can be added to the circuit to produce this effect without removing existing circuit elements? Must any existing connections be broken to accomplish this?
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