# Electrical Circuits

The basic equation governing the amount of current I (in amperes) in a simple RL circuit consisting of a resistance R (in ohms), an inductor L (in henries), and an electromotive force (abbreviated emf) E (in volts) is (dI)/(dt)+R/L I=E/L .
For an RC circuit consisting of a resistance, a capacitance C (in farads), an emf, and no inductance, the equation governing the amount of electrical charge q (in coulombs) on the capacitor is (dq)/(dt)+1/(RC) q=E/R .The relationship between q and I is I=(dq)/(dt) .

Example 1. An RL circuit has an emf of 5 volts, a resistance of 50 ohms, an inductance of 1 henry, and no initial current. Find the current in the circuit at any time t.

Here E=5, R=50, and L=1; hence differential equation becomes (dI)/(dt)+50I=5 . This is linear equation, its solution is I=ce^(-50t)+1/10 .

At t=0, I=0; thus, 0=ce^(-50*0)+1/10 or c=-1/10 . The current at any time, then, I=-1/10 e^(-50t)+1/10 .

The quantity -1/10 e^(-50t) is called the transient current, since this quantity goes to zero ("dies out") as t->oo .

The quantity 1/10 is called the steady-state current. As t->oo , the current I approaches the value of the steady-state current.

Example 2. A RC circuit has an emf of 300cos(2t) volts, a resistance of 150 ohms, a capacitance of 1/600 farad, and an

initial charge on the capacitor of 5 coulombs. Find the charge on the capacitor at any time t and the steady-state current.

Here E=300cos(2t) , R=150, C=1/600 .

So, (dq)/(dt)+4q=2cos(2t) . This is linear differential equation. It can be rewritten as (dq)/(dt)e^(4t)+4e^(4t)q=2e^(4t)cos(2t) or (d(qe^(4t)))/(dt)=2e^(4t)cos(2t) .

Integrating both sides gives qe^(4t)=int (2e^(4t)cos(2t))dt . Using integration by parts we obtain that qe^(4t)=c+(e^(4t))/5(sin(2t)+2cos(2t)) .

Charge on the capacitor at any time t is q=ce^(-4t)+1/5 (sin(2t)+2cos(2t)) .

Since q=5 when t=0 then 5=ce^(-4*0)+1/5(sin(2*0)+2*cos(2*0)) or c=23/5 .

Thus, q=1/5(23e^(-4t)+sin(2t)+2cos(2t)) .

Next since, I=(dq)/(dt) then I=1/5(-92e^(-4t)+2cos(2t)-4sin(2t)) and the steady-state current is I_s=1/5(2cos(2t)-4sin(2t)) (-92/5 e^(-4t) is transient because this quantity goes to zero as t->oo, unlike I_s ).