To solve the problem of the solenoid circuit, we need to break it down step by step. We have a solenoid with an inductance \( L = 1.8 \times 10^{-4} \) henry and a resistance \( R = 6 \) ohms. After dividing this solenoid into three identical pieces, we will analyze the circuit when two of these pieces are connected in parallel across a 12 V battery. Let's find the time constant and the steady-state current in this setup.
Step 1: Understanding the Circuit Configuration
When the solenoid is divided into three identical pieces, each piece will have the same inductance and resistance. The inductance of each piece can be calculated as follows:
- Inductance of each piece: \( L_{piece} = \frac{L}{3} = \frac{1.8 \times 10^{-4}}{3} = 6.0 \times 10^{-5} \) henry
- Resistance of each piece: \( R_{piece} = \frac{R}{3} = \frac{6}{3} = 2 \) ohms
Step 2: Analyzing the Parallel Connection
When two of these pieces are connected in parallel, the total inductance \( L_{total} \) and total resistance \( R_{total} \) of the circuit can be calculated. For inductors in parallel, the formula is:
Inductance in Parallel:
\( \frac{1}{L_{total}} = \frac{1}{L_{1}} + \frac{1}{L_{2}} \)
Since both inductors have the same inductance:
\( \frac{1}{L_{total}} = \frac{1}{6.0 \times 10^{-5}} + \frac{1}{6.0 \times 10^{-5}} = \frac{2}{6.0 \times 10^{-5}} \)
Thus, we find:
\( L_{total} = \frac{6.0 \times 10^{-5}}{2} = 3.0 \times 10^{-5} \) henry
Step 3: Total Resistance in Parallel
For resistors in parallel, the total resistance \( R_{total} \) is given by:
\( \frac{1}{R_{total}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \)
Substituting the values:
\( \frac{1}{R_{total}} = \frac{1}{2} + \frac{1}{2} = 1 \)
Thus, we find:
\( R_{total} = 1 \) ohm
Step 4: Calculating the Time Constant
The time constant \( \tau \) of an RL circuit is given by the formula:
\( \tau = \frac{L_{total}}{R_{total}} \)
Substituting the values we calculated:
\( \tau = \frac{3.0 \times 10^{-5}}{1} = 3.0 \times 10^{-5} \) seconds
Step 5: Finding the Steady-State Current
In steady state, the current \( I \) through the circuit can be calculated using Ohm's Law:
\( I = \frac{V}{R_{total}} \)
Substituting the values:
\( I = \frac{12}{1} = 12 \) A
Final Summary
In conclusion, after analyzing the circuit, we find:
- The time constant of the current in the circuit is \( 3.0 \times 10^{-5} \) seconds.
- The steady-state current through the battery is \( 12 \) A.
It seems there was a discrepancy in the steady-state current you provided. Based on the calculations, the correct steady-state current is 12 A, not 8 A. If you have any further questions or need clarification on any part of this explanation, feel free to ask!
