Chapter 30: Problem 53
A solenoid having an inductance of \(6.30 \mu \mathrm{H}\) is connected in series with a \(1.20 \mathrm{k} \Omega\) resistor. (a) If a \(14.0 \mathrm{~V}\) battery is connected across the pair, how long will it take for the current through the resistor to reach \(80.0 \%\) of its final value? (b) What is the current through the resistor at time \(t=1.0 \tau_{L} ?\)
Short Answer
Step by step solution
Understand the Problem for Part (a)
Calculate Final Current
Find the RL Circuit Time Constant
Determine Time for 80% Current
Problem Statement for Part (b)
Substitute \( t = \tau_L \) into the Current Equation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ohm's Law
- \( I = \frac{V}{R} \)
- \( I \) is the electric current in amperes (A)
- \( V \) is the voltage in volts (V)
- \( R \) is the resistance in ohms (Ω)
In RL circuits, this law is crucial for calculating the final steady-state current. In our example, with a 14.0 V battery and a 1.20 kΩ resistor, the final current \( I_f \) is calculated by dividing the voltage by the resistance, giving us approximately 0.0117 A.
Time Constant
- \( \tau_L = \frac{L}{R} \)
- \( L \) is the inductance in henrys (H)
- \( R \) is the resistance in ohms (Ω)
Inductance
The ability of the solenoid to store energy in a magnetic field as current passes through it is a key aspect to consider in RL circuits. High inductance in a circuit means it can oppose changes in current more effectively, which affects how quickly the circuit can reach a new steady-state current.
Exponential Decay
- \( I(t) = I_f \left(1 - e^{-t/\tau_L}\right) \)
- \( I_f \) is the final steady-state current
- \( e \) is the base of the natural logarithm
- \( t \) is the time elapsed
- \( \tau_L \) is the RL circuit time constant
Electric Current Calculation
- \( I(t) = I_f \left(1 - e^{-t/\tau_L}\right) \)
At \( t = \tau_L \), a typical checkpoint in RL circuits, the current reaches \( I(\tau_L) = I_f \left(1 - \frac{1}{e}\right) \), which calculates to approximately 0.0074 A when using the figures provided in the exercise. This shows that even at the time constant, the current is still rising towards its final stable value of about 0.0117 A.