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Chapter 19: Problem 91

A capacitor that is initially uncharged is connected in series with a resistor and a 400.0 \(\mathrm{V}\) emf source with negligible internal resistance. Just after the circuit is completed, the current through the resistor is 0.800 \(\mathrm{mA}\) and the time constant for the circuit is 6.00 s. What are (a) the resistance of the resistor and (b) the capacitance of the capacitor?

Short Answer

Expert verified
(a) 500 k惟; (b) 1.2 脳 10^-5 F

Step by step solution

01

Understanding the Problem

You've a series circuit with a capacitor initially uncharged, a resistor, and a 400 V emf connected. Just after closing the circuit, the current is 0.800 mA and the time constant is 6.00 s. You need to find the resistance and capacitance.
02

Using Ohm's Law

Use Ohm's Law to find the resistance. The initial current through the resistor is given by the formula \( I = \frac{V}{R} \). Rearrange to get \( R = \frac{V}{I} \). Substitute \( V = 400 \text{ V} \) and \( I = 0.800 \text{ mA} = 0.0008 \text{ A} \): \[ R = \frac{400}{0.0008} = 500000 \text{ ohms or } 500 \text{ k}\Omega. \]
03

Applying the Time Constant Formula

The time constant \( \tau \) of an RC circuit is given by \( \tau = RC \). Rearrange this to find the capacitance: \( C = \frac{\tau}{R} \). Substitute \( \tau = 6.00 \text{ s} \) and \( R = 500000 \Omega \): \[ C = \frac{6.00}{500000} = 1.2 \times 10^{-5} \text{ F} \].
04

Conclusion

After calculating, the resistance of the resistor is \( 500 \text{ k}\Omega \) and the capacitance of the capacitor is \( 1.2 \times 10^{-5} \text{ F} \).

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental concept in the study of electric circuits. It describes the relationship between voltage (V), current (I), and resistance (R). The law is represented by the formula:
  • \( V = IR \)
This formula shows that the voltage across a resistor is equal to the product of the current flowing through it and its resistance.
In context, if you know two of these values, you can easily find the third. For example, in the given exercise, you are provided with the voltage (400 V) and the current (0.800 mA). By rearranging the formula to solve for resistance, you use:
  • \( R = \frac{V}{I} \)
This allows us to determine that the resistance in the circuit is 500 k惟.
Time Constant
In an RC circuit, the time constant \( \tau \) is a crucial parameter that characterizes how quickly the circuit responds to changes. It is defined as the time it takes for the voltage across the capacitor to either charge up to 63.2% of its maximum value from zero (or discharge to 36.8% of its initial value if starting at full charge).
The formula for the time constant in an RC circuit is:
  • \( \tau = RC \)
where \( R \) is the resistance, and \( C \) is the capacitance of the circuit. In our exercise, we know that \( \tau = 6.00 \) seconds. Using this along with the resistance, we can find the capacitance using the rearranged form:
  • \( C = \frac{\tau}{R} \)
Finding this gives us a capacitance of \( 1.2 \times 10^{-5} \) F.
Capacitance
Capacitance is a measure of a capacitor's ability to store an electric charge. The unit of capacitance is the farad (F), and it explains how much charge a capacitor can store per volt of electric potential difference.
In simple terms, a capacitor with a high capacitance can store more charge than one with a lower capacitance when subjected to the same voltage. The capacitance is directly used in calculating the time constant of an RC circuit.
  • \( C = \frac{\tau}{R} \)
Using our findings from the exercise, the capacitance value of \( 1.2 \times 10^{-5} \) F indicates how effective this capacitor is in this particular circuit configuration.
Resistance
Resistance quantifies how strongly a component, like a resistor, opposes the flow of electric current. It is measured in ohms (惟) and is a central concept in circuit analysis.
In our context, resistance determines how much current will flow for a given voltage applied across the circuit. From Ohm's Law, we understand that:
  • \( R = \frac{V}{I} \)
Re-evaluating this formula with a 400 V source and a current of 0.800 mA, the resistance of the resistor is found to be 500 k惟. This tells us how difficult it is for current to pass through the resistor.
Series Circuit
A series circuit is a basic type of circuit in which components are connected end-to-end in a single path for current flow. In a series circuit, the same current flows through each component, but the total resistance is the sum of the individual resistances of each component.
This characteristic affects both the voltage distribution across components and the current flow throughout the circuit. For example, when a resistor and capacitor are placed in series, the voltage drop across each element will differ.
  • Voltage is divided: \( V_{total} = V_1 + V_2 + \ldots + V_n \)
  • Current is constant: \( I_{total} = I_1 = I_2 = \ldots = I_n \)
In our exercise, being a series circuit, it means that as the capacitor charges, the current is initially dictated by the total resistance.

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Most popular questions from this chapter

Charging and discharging a capacitor. A 1.50\(\mu \mathrm{F}\) capacitor is charged through a 125\(\Omega\) resistor and then discharged through the same resistor by short-circuiting the battery. While the capacitor is being charged, find (a) the time for the charge on its plates to reach \(1-1 / e\) of its maximum value and (b) the current in the circuit at that time. (c) During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1/e of its initial value. Also, find the time for the current in the circuit to decrease to 1\(/ e\) of its initial value.

Electricity through the body, II. The average bulk resistivity of the human body (apart from surface resistance of the skin) is about 5.0\(\Omega \cdot \mathrm{m} .\) The conducting path between the hands can be represented approximately as a cylinder 1.6 \(\mathrm{m}\) long and 0.10 \(\mathrm{m}\) in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of 100 \(\mathrm{mA}\) ? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?

A ductile metal wire has resistance \(R .\) What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched. (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Lightbulbs. The wattage rating of a lightbulb is the power it consumes when it is connected across a 120 potential difference. For example, a 60 W lightbulb consumes 60.0 \(\mathrm{W}\) of electrical power only when it is connected across a 120 \(\mathrm{V}\) potential difference. (a) What is the resistance of a 60 \(\mathrm{W}\) lightbulb? (b) Without doing any calculations, would you expect a 100 \(\mathrm{W}\) bulb to have more or less resistance than a 60 \(\mathrm{W}\) bulb? Calculate and find out.

Lightbulbs in series, I. The power rating of a lightbulb is the power it consumes when connected across a 120 \(\mathrm{V}\) out-let. (a) If you put two 100 \(\mathrm{W}\) bulbs in series across a 120 \(\mathrm{V}\) outlet, how much power would each consume if its resistance were constant? (b) How much power does each one consume if you connect them in parallel across a 120 \(\mathrm{V}\) outlet?
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