Transient

SAMPLE PROBLEMS: 112-Topic 9: INDUCTANCE & TRANSIENT DC CIRCUIT PROBLEMS: 1) A 2 nF capacitor is to be charged by placing it in series with a 1 M Ω resistor and connecting the circuit to the terminals of a 12 V battery. Determine: (a) the final charge on the capacitor; (b) the time constant of the circuit; (c) the initial current in the circuit; (d) the circuits 'half-life'; (d) the time required for the charge to reach 90% of its final value. Solution: The circuit is a shown. We close the switch at time t = 0. The time characteristics of the circuit are determined solely by the resistance and capacitance values. Thus we have:

i(t) S 12V

τ (time constant) = RC = (2 x 10-9)(1 x 106) = 2 m sec. 2 nf

T 1/2 (half-life) = .693 τ = 1.386 m sec.

1M Ω +

We analyze the circuit first from a 'physical' perspective. That is, let's ask ourselves what we expect to happen? The quantities of interest are q(t) the charge on the capacitor; vc(t) the potential difference across the capacitor; i(t) the current in the circuit; and vR(t) the potential difference across the resistor. We know the following: q(t): The charge on the capacitor starts (t=0) at zero, and builds up to a final value. Hence, the potential difference across C starts at 0 and builds up to a final value. i(t): The current in the circuit disappears when the capacitor is fully charged. Thus i(t) starts at some initial value Io and goes to 0. The potential difference across R behaves the same as i(t). Since i( ∞ )= 0, then vR( ∞ ) = 0, and the entire 12 V is across the capacitor. Thus: vc( ∞ ) = 12 V, and q( ∞ ) = Qf = Cvc( ∞ ) = 24 n C. Since q(t=0) = 0 then vc(0) = 0 and vR(0) = 12 V. Hence Io = i(t=0) = 12/R = 12 µ A. Making a table, we can now fill in the values of the physical quantities of interest at the four key times: 0, T1/2 , τ , and ∞ . variable q(t)

t=0

t = T(1/2)

0

(1/2)Qf = 12 nC

t=τ Qf(1 - 1/e) = .63 Qf = 15.2 nC

t=∞ 24 nC

i(t)

0 I0 = 12 A

(1/2) Vf = 6 V (1/2) I0 = 6 A

.63 Vf = 7.58 V (1/e) I0 = .37 I0 = 4.41 A

12 V 0

VR(t)

V0 = 12 V

(1/2) V0 = 6 V

(1/e) V0 = .37 V0 = 4.41 V

0

Vc(t)

The shape of the curves describing these variables can now plotted, and their corresponding mathematical functions can be written down:

q(t)

q(t) = Qf {1 - exp(- t/ τ )} = (24 nC){1 - exp[-t/(2x10-3)] f q(t)

Q

Time

i(t)

-3

i(t)

i(t) = Io exp(- t/τ ) = (12 A) exp[-t/(2x10 )]

Time

The variable vc(t) has the same shape as q(t) since vc(t) = (1/C) q(t), and vR(t) has the same shape as i(t) since VR(t) = R i(t) . Let us calculate (for practice) the values of these functions at some time other than the 4 key times. Consider time t' = 1 m sec. We would have: t'/τ = (1 msec)/(2 msec) = .5 and q(t') = (24 nC){1 - exp(-.5)} = (24 nC){1 -.6065} = (24 nC)(.3935) = 9.44 nC VC(t') = (1/C) q(t') = 4.72 V i(t') = Io exp(-.5) = (12 µ A)(.6065) = 7.278 A; V R(t') = R i(t') = (1 M Ω )(7.278 µ A) = 7.278 V We have calculated all desired quantities but one. We are asked for the time required for the charge to reach 90% of its final value. Calling this time t', we have: q(t') = .9 Qf = Qf {1 - exp(-t'/τ)} 

exp(-t'/τ) = .1

We first 'flip-over' this equation: 10 = exp(+t'/τ ). Now take the natural log of both sides: ln 10 = t'/τ  t' = (ln 10) τ . We need a calculator for this calculation. We find

ln 10 = 2.30  t' = (2.30) τ = 4.61 m sec.

2). A coil of self-inductance 0.5 H and resistance 100 ohms is connected across the terminals of a 12 V battery. If the connection is made at time t = 0, determine: (a) the initial current in the circuit; (b) the final current; (c) the 'half-life' of the circuit; (d) the current and back emf values at .005 seconds after the connections are made. Solution: The circuit is a shown. We close the switch at time t = 0. The time characteristics of the circuit are determined solely by the inductance and resistance values. Thus we have: τ (time constant) = L/R = (.5)/(100) = 5 m sec. T1/2 (half-life) = .693

S 12 V 100

.5 h Ω

= 3.47 m sec.

We analyze the circuit first from a 'physical' perspective. That is, let's ask ourselves what we expect to happen? The quantities of interest are i(t) the current in the circuit; and vR(t) the potential difference across the resistor; the back emf in the coil, E = L di/dt. (If we wish, we can include the 'rate of change of the current' as a 4th variable. We know the following: the final state (steady state) situation, is that there is a (constant) dc current If in the circuit, and the back emf of the coil has disappeared. Hence: E (t): The back emf starts (t=0) at Vo, and decreases to a final value of zero. This is the same as vL(t), the potential difference across L. i(t): The current in the circuit starts at 0 and builds up to a final value I f. The potential difference across R behaves the same as i(t). di/dt: Behaves the same as the back emf (they are proportional). Thus this starts at a maximum value and decreases to zero. Since i( ∞ )= If, then vR( ∞ ) = IfR = the entire 12 V is across the resisitor. Thus: If = 12/100 = .12 A. Since i(t=0) = 0 then vR(0) = 0 and vL(0) = 12 V. Hence di/dt @ t=0 = 12/L = 24 A/sec. Making a table, we can now fill in the values of the physical quantities of interest at the four key times: 0, T1/2 , τ , and ∞ .

(1/2)I f = .06 A

t=τ I f (1 - 1/e) = .63 If = .076 A

t=∞ .12 A

di/dt

0 24 A/s

(1/2) Vf = 6 V (1/2) 24 = 12 A/s

.63 Vf = 7.58 V (1/e) 24 = .37 (24) = 8.83 A/s

12 V 0

E(t)

V0 = 12 V

(1/2) V0 = 6 V

(1/e) V0 = .37 V0 = 4.41 V

0

variable i(t) VR(t)

t=0

t = T(1/2)

0

The shape of the curves describing these variables can now plotted, and their corresponding mathematical functions can be written down:

E(t)

E(t)

E (t) = Vo exp(- t/ τ ) = (12 V) exp[-t/(5x10-3)]

Time

i(t)

i(t) = If {1 - exp(- t/τ )} = (.12 A){1 - exp[-t/(5x10-3)] f i(t)

I

Time

The variable di/dt(t) has the same shape as E(t) since di/dt = E /L, and vR(t) has the same shape as i(t) since VR(t) = R i(t) . Let us calculate (for practice) the values of these functions at some time other than the 4 key times. Consider time t' = 1 m sec. We would have: t'/τ = (1 msec)/(5 msec) = .2 and i(t') = (.12 A){1 - exp(-.2)} = (.12 A){1 -.8187} = (.12 A)(.1813) = .022 A; E (t') = Vo exp(-.2) = (12 V)(.8187) = 9.83 V;

VR(t') = R i(t') = 2.2 V

{di/dt}(t') = E(t')/L = (9.83 )/(.5) = 19.7 A/sec

We have calculated all desired quantities. Note that we are asked for the values of the current and emf at time t= 5 msec after the switch is closed. It just happens that this particular time is, in fact, the time constant L/R. Hence, the answers are given in the table above. 3) A 2µf capacitor is charged by attaching it to a 12 Volt supply. After becoming fully charged the capacitor is then connected through a switch to a 10 Ω resistor. If the switch is closed at time t = 0, determine: a) the time constant and half-life of the circuit b) The initial current in the circuit, c) the current in the circuit when the charge on the capacitor is 1/4th its initial value.

Solution: The circuit is a shown. We close the switch at time t = 0. The time characteristics of the circuit are determined solely by the resistance and capacitance values. Thus we have:

i(t) S +

τ (time constant) = RC = (2 x 10-6)(10) = 20 µ sec.

2 µf

10 Ω

T 1/2 (half-life) = .693 τ = 13.86 µ sec. We analyze the circuit first from a 'physical' perspective. That is, let's ask ourselves what we expect to happen? The quantities of interest are q(t) the charge on the capacitor; vc(t) the potential difference across the capacitor; i(t) the current in the circuit; and vR(t) the potential difference across the resistor. We know the following: q(t): The charge on the capacitor starts (t=0) at Q0, and drops to a final value of zero. Hence, the potential difference across C starts at V0 and drops to a final value of zero. i(t): The current in the circuit disappears when the capacitor is fully charged. Thus i(t) starts at some initial value Io and goes to 0. The potential difference across R behaves the same as i(t). Since the voltage across the capacitor is equal to the voltage across the resistor at all times, we have: Q 0 = C V0 = (2 µf)(12 V) = 24 µC. The initial current is then: I0 = V0/R = 1.2 A. Making a table, we can now fill in the values of the physical quantities of interest at the four key times: 0, T1/2 , τ , and ∞ . variable q(t)

t=0

t = T(1/2)

i(t)

Q0 = 24 µC V0 = 12 V I0 = 1.2 A

VR(t)

V0 = 12 V

Vc(t)

(1/2)Q0 = 12 µC (1/2) V0 = 6 V (1/2) I0 = .6 A

t=τ (1/e)Q0 = .37 Q0 = 8.88 µC (1/e) V0 = .37 V0 = 4.44 V (1/e) I0 = .37 I0 = .441 A

t=∞ 0

(1/2) V0 = 6 V

(1/e) V0 = .37 V0 = .441 V

0

0 0

The shape of the curves describing these variables can now plotted, and their corresponding mathematical functions can be written down: q(t)

q(t)

q(t) = Qo exp(- t/ τ )} = (24 nC) exp[-t/(20x10-6)]

Time

i(t)

-6

i(t)

i(t) = Io exp(- t/τ ) = (1.2 A) exp[-t/(20x10 )]

Time

The variable vc(t) has the same shape as q(t) since vc(t) = (1/C) q(t), and vR(t) has the same shape as i(t) since VR(t) = R i(t) . Let us calculate (for practice) the values of these functions at some time other than the 4 key times. Consider time t' = 10 µ sec. We would have: t'/τ = (10 µsec)/(20 µsec) = .5 and q(t') = (24 nC) exp(-.5)} = (24 nC)(.6065) = 14.556 µC; i(t') = Io exp(-.5) = (1.2 µ A)(.6065) = .7278 A; V

VC(t') = (1/C) q(t') = 7.25 V

V R(t') = R i(t') = (10 Ω )(.7278 µ A) = 7.278

In order to calculate the current for a given value of the charge we turn to the Differential Equation. Once the switch is closed, the potential differences across the capacitor and resistor are the same: VR = VC  i R = (1/C) q . Since the capacitor is discharging we have: i = - dq/dt . Hence: dq/dt = - (1/RC) q . This holds for all times. Hence, when the current is 1/4th its original value (6 µC), the current is [1/(20 µsec)] (6 µC) = .3 A . We can also determine this result another way. Since 1/4 = (1/2)(1/2) then the time at which this occurs is twice the half-life. Hence, the current is also 1/4th of its original value. 4) A solenoid consists of 100 turns, each or area 10 cm2. The length of the solenoid is 15 cm. What is the self-inductance of this solenoid? Solution: The field of a solenoid is B = µ o n I where 'n' is the number of turns per unit length. That is: n = N/L . If the current is varying in the solenoid, then an induced Emf occurs. From Faraday's Law we have: E = dΦ/dt Now the flux through the solenoid is: Φ = N B A = N µ o n A I . (We write the current as ' i ' since it is time varying.) Thus: E = N µ o n A di/dt and the self-inductance is L = µ o N2 A/L = 4π x 10-7 (100)2 (10-2)/.15 =

8.38 x 10-5 H.

5) A 20 micro farad capacitor is charged by attaching it to a 200 V battery. The capacitor is then disconnected from the battery and is connected through a switch to a 5000 ohm resistor. a) Sketch the form of the charge on the capacitor and the current in the circuit as functions of time after the switch is closed. b) Determine the maximum current. c) At what time does the charge drop to one half of its original value? Solution: The circuit is a shown. We close the switch at time t = 0. The time characteristics of the circuit are determined solely by the resistance and capacitance values. Thus we have:

i(t)

+

τ (time constant) = RC = (5 x 103)(2 x 10-6) = 10 x 10 -3 = .01 sec

20 µ f

S 5k Ω

T 1/2 (half-life) = .693 τ = 6.7 m sec. (Ans. c) The initial charge on the capacitor is Q 0 = CV = (20 µ)(200) = 4000 µC = 4 mC. At the instant the switch is thrown the voltage across the resistor is 200 Volts. Hence, I 0 = 200/5000 = .04 A .(Ans. b) q(t)

q(t)

q(t) = Qo exp(- t/ τ )} = (4 mC) exp[-t/.01]

Time

i(t)

i(t)

i(t) = Io exp(- t/τ ) = (0.04 A) exp[-t/.01]

Time