Capacitors and Inductors with Current and Voltage Represented Analytically

Introduction

Each of the circuits in this problem set contains of a single capacitor or inductor and a single independent source. When the independent source is a current source, the current source current is equal to the current in the capacitor or inductor. Similarly, when the independent source is a voltage source, the voltage source voltage is equal to the voltage across the capacitor or inductor.

These problems can be solved using the constitutive equations for the capacitor and inductor. (An “constitutive equation” is the equation that describes the relationship between the element voltage and element current.) Capacitors and inductors are described in Sections 7.2 and 7.5 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda.

Worked Examples

Example 1: The input to the circuit shown in Figure 1 is the current

it()= 3.75 e−1.2t A for t> 0

The output is the capacitor voltage

vt()= 4−> 1.25 e−1.2t V for t 0

Find the value of the capacitance, C.

Figure 1 The circuit considered in Example 1.

Solution: The capacitor voltage is related to the capacitor current by

2 1 t vt()=+ i()ττ d v ()0 C ∫0 That is

1 t 3.75 t −3.125 4− 1.25eedveve−−1.2t = 3.75 1.2τττ +=()0 −1.2 +=()0 ()−1.2t −+1v() 0 CC∫0 ()−1.2 0 C

Equating the coefficients of e−1.2t gives

3.125 3.125 12.5 =⇒===C 0.25 250 mF C 12.5

Example 2: The input to the circuit shown in Figure 2 is the current

it()= 3−> 4.5 e−6t A for t 0

Determine the inductor voltage,vt(), for t > 0.

Figure 2 The circuit considered in Example 2.

Solution: The inductor voltage is related to the inductor current by

dd vt()==−=−−= L it() 2.5() 3 4.5e−−66tt 2.5() 4.5() 6e 67.5 e−6t V dt dt

Example 3: The input to the circuit shown in Figure 3 is the voltage

vt( ) =15 e−4t V for t> 0

3 The initial current in the inductor is i(02) = A. Determine the inductor current,it(), for t > 0.

Figure 3 The circuit considered in Example 3.

Solution: The inductor current is related to the inductor voltage by

1 t it()=+ v()ττ d i ()0 L ∫0 That is

11t 5t it()=+15 e−−−−4444ττ dτ 2 =+e 2=−− 1.5()e tt 1+ 2=− 3.5 1.5e A for t> 0 2.5 ∫0 2.5()− 4 0

Example 4: The input to the circuit shown in Figure 4 is the voltage

vt()= −+6 12 e−5t V for t> 0

The output is the capacitor current

it()= −>4 e−5t mA for t 0

Find the value of the capacitance, C.

Figure 4 The circuit considered in Example 4.

4 Solution: The capacitor current is related to the capacitor voltage by

d it()= C vt() dt That is d −=−+=−=−0.004eC−−−555ttt()6 12 e C() 12() 5 e 60 Ce−5t V dt

Equating the coefficients of e−5t gives

0.004 −=−⇒==×=0.004 60CC 6.67 10−5 F 66.7 μ F 60

Example 5: The input to the circuit shown in Figure 5 is the current

it( ) = 3 e−25t A for t> 0

The initial capacitor voltage is vC ()02=− V. Determine the current source voltage, vt(), for t > 0 .

Figure 5 The circuit considered in Example 5.

Solution: Apply KVL to the mesh to get

⎡ 1 t ⎤ vt()=+=+44 it () vC () t it () i()ττ d + v ()0 ⎣⎢0.05 ∫0 ⎦⎥ That is,

13t vt()=+43 e−−−25tt3e25τ dτ −=+ 2 12 e25 e −25t −−1 2 () ∫0 () 0.05 0.05()− 25 =12ee−−25tt−−−=+ 2.4()25 1 2 9.6 e −25 t 0.4 V for t> 0

5 Example 6: The input to the circuit shown in Figure 6 is the current

it( ) = 3 e−25t A for t> 0 The output is the voltage

vt()= 9.6 e−25t +> 0.4 V for t 0

The initial capacitor voltage is vC ()02=− V. Determine the values of the capacitance, C, and resistance, R.

Figure 6 The circuit considered in Example 6.

Solution: Apply KVL to the mesh to get

⎡ 1 t ⎤ vt()=+=+ Rit () vC () t Rit () i()ττ d + v ()0 ⎣⎢C ∫0 ⎦⎥ That is

⎡⎤1 t 9.6eReed−−−25tt+= 0.4() 3 25 +325τ τ − 2 ⎣⎦⎢⎥C ∫0

−−25tt3125 ⎛⎞−25t3 =+31Re ()e −−=−+23⎜⎟R e −2 CC()−25 ⎝⎠25 25C

Equating coefficients gives 3 0.4=−⇒== 2C 0.05 50 mF 25C and ⎛⎞11⎛⎞

9.6=− 3⎜⎟RR =−3⎜⎟ =−3() RR 0.8 ⇒= 4 Ω ⎝⎠25C ⎝⎠25() 0.05

6 Example 7: The input to the circuit shown in Figure 7 is the voltage

vt( ) = 4 e−20t V for t> 0 The output is the current it()= −−1.2 e−20t 1.5 A for t> 0

The initial inductor current is iL ()0=− 3.5 A . Determine the values of the inductance, L, and resistance, R.

Figure 7 The circuit considered in Example 7.

Solution: Apply KCL at either node to get

vt() vt( ) ⎡ 1 t ⎤ it()=+ iL () t =+v()ττ d + i ()0 RRL⎣⎢ ∫0 ⎦⎥ That is

−−20tt20 41eet 4 4 −−=+1.2ee−−20t 1.5 420τ dτ −=+ 3.5 () e −20t −−1 3.5 RL∫0 RL()−20

⎛⎞41−20t 1 =−⎜⎟e +−3.5 ⎝⎠RL55 L Equating coefficients gives 1 −=1.5 − 3.5 ⇒L = 0.1 H 5 L and 41 4 1 4 −1.2 =− =− =−2 ⇒R =Ω 5 RLR550.1() R

Example 8: The input to the circuit shown in Figure 8 is the voltage

vt()= 8+> 5 e−10t V for t 0

Determine the current, it() for t > 0 .

Figure 8 The circuit considered in Example 8.

Solution: Apply KCL at either node to get

vt() ded85+ −10t it()=+0.05 vt() = +0.05() 8 + 5e−10t 44dt dt =+2 1.25eee−−−10ttt + 0.05()( 5 − 10 ) 10 =− 2 1.2510 A for t> 0