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Chapter 26: Problem 14

A battery, a resistor, and a capacitor are connected in series in an RC circuit. What happens to the current through a resistor after a long time? Explain using Kirchhoff's rules.

Short Answer

Expert verified
The behavior of the current through the resistor is that it becomes zero after a long time as the capacitor becomes fully charged. This is explained using Kirchhoff's Voltage Law (KVL) and Ohm's Law.

Step by step solution

01

Understanding the RC Circuit Components

An RC circuit consists of a resistor (R), a capacitor (C), and a battery. The resistor provides resistance to the current flow, the capacitor stores energy in an electric field, and the battery provides a constant voltage (V) across the circuit.
02

Understanding Kirchhoff's Rules

Kirchhoff's rules consist of two rules that help us understand and analyze electric circuits. The first rule, Kirchhoff's Current Law (KCL), states that the total current entering a junction in a circuit must equal the total current leaving the junction. The second rule, Kirchhoff's Voltage Law (KVL), states that the sum of the voltage drops across all elements in a closed loop must be equal to the voltage supplied by the battery.
03

Applying Kirchhoff's Voltage Law (KVL) to the RC Circuit

In an RC circuit, we can set up a loop consisting of the battery, the resistor, and the capacitor. According to KVL, the sum of the voltage drops across the resistor (V_R) and the capacitor (V_C) should be equal to the battery voltage (V). V = V_R + V_C
04

Calculating the Voltage Drop across the Resistor (V_R)

The voltage drop across the resistor can be calculated using Ohm's Law, which states that the voltage across a resistor is directly proportional to the current flowing through it (I) multiplied by the resistance (R). V_R = I * R
05

Calculating the Voltage Drop across the Capacitor (V_C)

The voltage across a capacitor can be expressed as: V_C = Q / C Where Q is the charge stored on the capacitor.
06

Determining the Charge on the Capacitor after a Long Time

After a long time, the capacitor becomes fully charged, and the voltage across it (V_C) will be equal to the battery voltage (V). At this point, the capacitor acts as an open circuit, so no current will flow through it.
07

Calculating the Current through the Resistor after a Long Time

Since the voltage across the capacitor (V_C) is equal to the battery voltage (V) after a long time, there will be no voltage drop across the resistor (V_R). Therefore, according to Ohm's Law (V_R = I * R), the current (I) through the resistor will be zero. In conclusion, after a long time, the current through the resistor becomes zero as the capacitor becomes fully charged. This behavior is explained using Kirchhoff's Voltage Law (KVL) and Ohm's Law.

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Most popular questions from this chapter

Consider a series \(\mathrm{RC}\) circuit with \(R=10.0 \Omega\) \(C=10.0 \mu \mathrm{F}\) and \(V=10.0 \mathrm{~V}\) a) How much time, expressed as a multiple of the time constant, does it take for the capacitor to be charged to half of its maximum value? b) At this instant, what is the ratio of the energy stored in the capacitor to its maximum possible value? c) Now suppose the capacitor is fully charged. At time \(t=\) 0 , the original circuit is opened and the capacitor is allowed to discharge across another resistor, \(R^{\prime}=1.00 \Omega\), that is connected across the capacitor. What is the time constant for the discharging of the capacitor? d) How many seconds does it take for the capacitor to discharge half of its maximum stored charge, \(Q\) ?

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) \(4 \tau\)

Explain why the time constant for an \(\mathrm{RC}\) circuit increases with \(R\) and with \(C\). (The answer "That's what the formula says" is not sufficient.)

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.)

A \(12.0-V\) battery is attached to a \(2.00-\mathrm{mF}\) capacitor and a \(100 .-\Omega\) resistor. Once the capacitor is fully charged, what is the energy stored in it? What is the energy dissipated as heat by the resistor as the capacitor is charging?
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