58.5k views
5 votes
A50 V battery is connected to a solenoid with inductance 45 H using a wire with a resistance of 9 ohms Calculate the time it takes for the current to build up to 2 Amos this time bigger than less than or equal to the time constant of the circuit? Explain why, based on how the time constant is definest, and without just using the formula for the time constant

A50 V battery is connected to a solenoid with inductance 45 H using a wire with a-example-1
User ZooKeeper
by
8.2k points

1 Answer

5 votes

The time constant of the circuit is the product of the resistance and the inductance, and it represents the time it takes for the current to reach 63.2% of its maximum value when changing from 0 to its final value.

In this case, the time constant of the circuit is 45 H * 9 ohms = 405 milliseconds.

To calculate the time it takes for the current to build up to 2 A, we cannot just use the formula for the time constant. Instead, we need to use the exponential equation for the current in an RL circuit:

i(t) = i_final * (1 - e^(-t/τ))

where τ is the time constant.

Setting i(t) = 2 A, we can solve for t:

2 = 2 * (1 - e^(-t/405 ms))

e^(-t/405 ms) = 0.5

-t/405 ms = ln(0.5)

t = 405 ms * ln(2)

The time it takes for the current to build up to 2 A is less than the time constant of the circuit. This means that the circuit will reach 63.2% of its maximum current within 405 ms, but it will take longer to reach 2 A.

User Mongmong Seesee
by
7.0k points