Applications of Differential Equations in Electrical Circuits

Question

1. A series RL circuit with R=40Ω and L=1.5H has a constant voltage E=5V applied at t=0 by the closing of switch. Find the equation of current as a function of time.
2. A series RL circuit with R=40Ω and L=1.5H has a constant voltage E=5V applied at t=0 by the closing of switch. Find V_L(t).
3. A series RL circuit with R=40Ω and L=1.5H has a constant voltage E=5V applied at t=0 by the closing of switch. Find V_R(t).

NOTE: Show your detailed solutions.

Answer 1

For these kind of problems you need to build the differential equation for the circuit, remembering three thigs:

1. Currents in series are equal
2. Voltage drops across resistors are V_R = R I
3. Voltage drops across Inductors are V_L = L ∂_tI

So by using the loop rule on the loop containing the battery, resistor and inductor we get the differential equation:

5V - 40Ω I - 1.5H ∂_tI = 0

To solve the equation we can define I_p = I - 5V/40Ω = I - 1/8 A, with this new variable the differential equation is:

- 40Ω I_p - ∂_t I_p 1.5H = 0

Which can be solved to obtain:

I_p(t) = I_p(0) e^{-80/3 Hz t}

since at t=0 I(0) = 0 this means that I_p(0) = -1/8 A. Then by solving for I we get:

I(t) = I_p(t) + 1/8 A = 1/8A*(1-e^{-80/3Hz t} )

And the Voltages are:

V_R(t) = R I(t) = 5V * (1-e^{-80/3Hz t} )

V_L(t) = L ∂_tI(t) = 1.5H*1/8A*(-80/3Hz)*e^{-80/3Hz t} = -5V e^{-80/3Hz t}