A uniform wound solenoid of L= 1.8 * 10^-4 henry and R=6 ohm is broken up into 3 identical piece

these 2 coils are then connected in parallal acrossa 12 V battery

The time constant of the current in the circuit and the steady state current through battery is

Ans 3*10^-5 , 8A

Please give me a detailed solution i will ofcourse rate you

To solve the problem of the solenoid's current characteristics when broken into three identical pieces and connected in parallel, we need to analyze the circuit step by step. Let's break this down into manageable parts: calculating the time constant and the steady-state current.

Understanding the Components

We start with a solenoid that has an inductance (L) of 1.8 x 10^-4 henries and a resistance (R) of 6 ohms. When this solenoid is divided into three identical coils, each piece will have the same inductance and resistance characteristics.

Inductance and Resistance of Each Coil

• Inductance of each piece: Since the solenoid is uniform and divided into three parts, the inductance of each coil will be:

L_coil = L / 3 = (1.8 x 10^-4 H) / 3 = 6 x 10^-5 H

• Resistance of each piece: The resistance remains the same for each coil since they are identical:

R_coil = R = 6 ohms

Connecting Coils in Parallel

When the two coils are connected in parallel, the total inductance (L_total) and total resistance (R_total) of the circuit can be calculated using the following formulas:

Total Inductance in Parallel

The formula for inductance in parallel is given by:

1 / L_total = 1 / L_coil1 + 1 / L_coil2

Since both coils are identical:

1 / L_total = 1 / (6 x 10^-5 H) + 1 / (6 x 10^-5 H) = 2 / (6 x 10^-5 H)

Thus, we find:

L_total = (6 x 10^-5 H) / 2 = 3 x 10^-5 H

Total Resistance in Parallel

The formula for resistance in parallel is:

1 / R_total = 1 / R_coil1 + 1 / R_coil2

Again, since both resistances are equal:

1 / R_total = 1 / 6 + 1 / 6 = 2 / 6

This simplifies to:

R_total = 6 / 2 = 3 ohms

Calculating the Time Constant

The time constant (τ) of an RL circuit is given by the formula:

τ = L_total / R_total

Substituting the values we calculated:

τ = (3 x 10^-5 H) / 3 ohms = 1 x 10^-5 seconds

Finding the Steady-State Current

The steady-state current (I) through the battery can be calculated using Ohm's Law:

I = V / R_total

Where V is the voltage of the battery:

I = 12 V / 3 ohms = 4 A

Final Results

To summarize, the time constant of the current in the circuit is:

τ = 1 x 10^-5 seconds

And the steady-state current through the battery is:

I = 4 A

It seems there was a misunderstanding in the initial answer you provided regarding the time constant and the steady-state current. The calculations show that the time constant is 1 x 10^-5 seconds and the steady-state current is 4 A. If you have further questions or need clarification on any part of this explanation, feel free to ask!