Problem Solving for solenoid with given inductance and resistance

Question

A solenoid has an inductance of 100mH and resistance R of 5.5Ω

How can I find the:

a) The time constant for this circuit

b) How long would it take for the current to go from 0 to 63% of its final maximum value when connected to a battery of voltage V?

c) Assuming that is then again connected to another power source, determine the current in the coil if

i)120V_dc is applied to it,

ii)120V_ac (rms) at 60Hz is applied

d) Supposedly, if we have we have a capacitance of 15 μF, and using the same inductance of L=100mH and resistance 5.5Ω and connecting them to the same AC power source 120 V (V_RMS = 120V_ac) running at 500 Hz, how will you solve for the current that is formed in the circuit when the power is on?

Answer 1

a) The time constant for an solenoid is given by dividing the inductance by the resistance:

τ=L/R = 100 mH / 5.5 Ω ≈ 18 sec

b) By definition, the time constant represents how long the current in a solenoid would take to reach 63.2% of its final maximum value, so about 18 sec.

c)

i) The current in the solenoid can be modeled as an LR circuit. According to Kirchhoff's loop law, the supplied potential of the circuit is equal to the drop of potential across the inductor and the resistor, which can be modeled separately:

V = IR + L (dI/dt)

This is a first order linear non-homogenous ordinary differential equation, so we can use the integrating factor method to find the solution:

I(t) = (V/R)[1-e^Rt/L]

ii) We can set this problem up the same way with Kirchhoff's loop law again, only replacing the constant supplied voltage with a time-varying one, as indicated.

V(t) = IR + L (dI/dt)

The only difference in this approach is that the first order linear non-homogeneous ordinary differential equation is now non-autonomous. We can still use the integrating factor method to find the solution:

I(t) = [(120 V) / (60 Hz)² / [(100 mH)² + (5.5 Ω)²]] · [(60 Hz) · (100 mH) e^-Rt/L - (60 Hz) ·(100 mH) · cos((60 Hz) · t) + (5.5 Ω) · sin((60 Hz) · t)]

d) This problem is what is called an LRC circuit, and it is set up the same way as the first two differential equations, although there is one subtlety here. The ODE must be represented in terms of the charge rather than the current. The current can then be solved for after solving the ODE by differentiating the result.

According to Kirchhoff's loop law, we have:

V(t) = IR + L (dI/dt) + (Q/C)

Since I = dQ/dt, we have

V(t) = (dQ/dt) · R + L · (d²Q/dt²) + (Q/C)

Note this is a second order linear non-homogeneous non-autonomous ordinary differential equation. We can use the method of undetermined coefficients, variation of parameters, or Laplace transforms in order to solve this problem. The corresponding initial value problem can be solved by using the initial conditions Q(t=0) = 0 and dQ/dt = I(t=0) = 0, as there is no charge or current before the voltage is supplied to the circuit.

I leave the solution as an exercise in differential equations. Feel free to reach out for help.