3.

4 Electrical Circuits:

L
I

E + C
−

I
R
An RLC circuit.

E is the voltage source (voltage drop in V, volts)

R is the resistor (resistance in Ω, ohms)

C is the capacitor (capacitance in F, farads)

L is the inductor (inductance in H, henrys)

I is the current (in A, amperes)

Component laws:

1. Ohm’s law: ER = RI
dI
2. Faraday’s law: EL = L
dt
Q
3. Capacitance law: EC = Q is the charge on the capacitor (in C, coulombs)
C
4. Kirchhoff’s voltage law: The sum of the voltage drops around any closed loop in a circuit
must be zero.

5. Kirchoff’s current law: The sum of currents flowing into any junction is zero.

From the law 4 we get EL + EC + ER − E = 0 or

1
 dI Q
L + + RI = E (1)
dt C
dQ
We use I = and differentiate (1) to get
dt

d2 I 1 dQ dI dE
L 2
+ +R = (2)
dt C dt dt dt
or
d2 I dI 1 dE
L 2
+R + I= (3)
dt dt C dt

If there is no capacitor then from (1) we obtain

dI
L + RI = E (4)
dt
Q
If there is no inductor then RI + = E or
C

dQ Q
R + =E (5)
dt C

1
Example 1: R= 2
Ω, L = 1H, E = 1 V and there is no capacitor. If I(0) = 0 then find I .
dI 1
Solution: The formula (4) gives the IVP + I = 1, I(0) = 0.
dt 2
It is a first order linear and separable equation. Its solution is I(t) = 2(1 − e−t/2 )

1
Example 2: R = 2 Ω, C = 5
F, E = cos t V and there is no inductor. Find I if I(0) = 0.
dQ
Solution: The formula (5) gives the IVP 2 + 5Q = cos t, I(0) = 0.
dt
5 5
R
It is a first order linear equation. The integrating factor is u(t) = e 2 dt = e 2 t
 5 0 5 5 5
Then e 2 t Q = 12 e 2 t cos t ⇒ e 2 t Q = 21 e 2 t (2 sin t + 5 cos t) + C ⇒

5 dQ 5
Q(t) = sin t + 52 cos t + C1 e− 2 t ⇒ I(t) = = cos t − 52 sin t + Ce− 2 t
dt
I(0) = 1 + C = 0 ⇒ C = −1
5
I(t) = cos t − 52 sin t − e− 2 t

2
• Exercise #10, page 131. An inductor (1H) and resistor (0.1Ω) are joined in series with
an electromotive force (emf) E = E(t) as in the Figure.

L
I

E +
−

I
R

If there is no current in the circuit at time t = 0, find the ensuing current in the circuit
at time t for the emf E(t) = 4 cos(3t) V .
Solution: L = 1H (Henry), R = 0.1Ω,
dI 1
E = ER + EL + 
E
C = RI + L + Q,
dt C
hence
dI
4 cos(3t) = 0.1I + 1 ,
dt
I 0 = −0.1I + 4 cos(3t), (·e0.1t )
Z Z
(Ie0.1t )0 = 4 cos(3t) · e0.1t , Ie0.1t = 4 cos(3t) · e0.1t dt, I(t) = 4e−0.1t cos(3t) · e0.1t dt.

Using integration by parts twice

4
10 cos(3t) + 300 sin(3t) + Ce−0.1t ,


I(t) =
901
and using the initial condition I(0) = 0 we obtain
40
C=− ,
901
hence
4
40 −0.1t
I(t) = 10 cos(3t) + 300 sin(3t) − e .
901 901
• Exercise #15, page 131. A resistor (20Ω) and capacitor (1F ) are linked in series with an
electromotive force (emf) E = E(t) in RC circuit.

3
 R
I

E + C
−

If the emf is given as E(t) = 10e−0.05t and the current is zero at time t = 0, find the
maximum charge on the capacitor and the time that it will occur.
Solution: R = 20Ω, C = 1F, Q(0) = 0,
1 1
E = ER + EC ≡ RI + Q = RQ0 + Q,
C C
t 1 1 t
10e− 100 = 20Q0 + Q, Q0 = − Q + e− 100
20 2
t
and using as integrating factor u(t) = e 20 we obtain
t
1 t
e 20 Q = e 25 ,
2
t 25 t
e 20 Q(t) = e 25 + C,
2
t t t t
Q(t) = 12.5e 25 − 20 + Ce− 20 ≡ 12.5e−0.01t + Ce− 20 .

Using the initial condition Q(0) = 0 ≡ 12.5 + C we obtain C = −12.5, and therefore

Q(t) = 12.5e−0.01t e−0.04t − 1 .

To find the maximum charge, we compute

Q0 (t) = e−0.05t 625 × 10−4 − 125 × 10−4 e0.04t ,

hence the maximum charge is when Q0 (T ) = 0, i.e.,

T = 25 ln(5),
Q(25 ln(5)) ≈ 6.6874.