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` ).