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# Solutions of Homework problems - PowerPoint PPT Presentation

Solutions of Homework problems. Resistive circuits. Problem 1 Use KVL and Ohms law to compute voltages v a and v b . -. v 1. From Ohms law: v 1 =8k W* i 1 =8[V]. v 2 =2k W* i 2 =-2[V] Form KVL: v a =5[V]-v 2 =7[V] v b =15[V]-v 1 -v a =0[V]. +. -. v 2. +. +. +. -. -.

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### Solutions of Homework problems

Problem 1

Use KVL and Ohms law to compute voltages va and vb .

-

v1

From Ohms law:

v1=8kW*i1=8[V]

v2=2kW*i2=-2[V]

Form KVL:

va=5[V]-v2=7[V]

vb=15[V]-v1-va=0[V]

+

-

v2

+

+

+

-

-

Problem 2

Write equations to compute voltages v1 and v2 , next find the current value of i1

From KCL:

50 mA=v1/40+(v1-v2)/40

and

100 mA=v2/80+(v2-v1)/40

i1

i1

v1

v2

Multiply first equation by 40:

2=v1+v1-v2=2v1-v2

From second equation:

10=2v2 => v2=5 [V], v1=1+v2 /2=3.5[V]

i1= (v1-v2)/40=-1.5/40=37.5 [mA]

100 mA

50 mA

Problem 3: Find Thevenin and Norton equivalent circuit for the network shown.

I1

N1

N2

I2

vt

From KVL

I1

N1

N2

Isc

I2

From KVL

_

Thevenin & Norton

RTh=-1.33Ω

A

A

RTh=vt/Isc=-1.33Ω

RTh=-1.33Ω

In=4.5 A

Vt=-6 V

Note: Negative vt indicates that the polarity is reversed and as a result this circuit has a negative resistance.

B

B

Norton Equivalent

Thevenin Equivalent

Problem 4: Find the current i and the voltage v across LED diode in the circuit shown on Fig. a) assuming that the diode characteristic is shown on Fig. b).

Draw load line. Intersection of load line and diode characteristic is the i and v across LED diode: v ≈ 1.02 V and i ≈ 7.5 mA.

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

i

+

v

_

(a)

2kΩ

Diode is on for v > 0 and R=2kΩ.

In a series connection voltages are added for each constant current

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

i

+

v

_

(a)

2kΩ

Resulting characteristics

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

i

1kΩ

+

v

_

(b)

+

_

Due to the presence of the 5V supply the diode conducts only for v > 5, R = 1kΩ

5V

First combine diode and resistance then add the voltage source

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

+

v

_

2kΩ

1kΩ

A

B

(c)

Diode B is on for v > 0 and R=1kΩ.

Diode A is on for v < 0 and R=2kΩ.

Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.

(d)

i

Diode D is on for v > 0 and R=1kΩ.

Diode C is on for v < 0 and R=0Ω.

+

v

_

D

C

1kΩ

_

+

_

Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.

1kΩ

+

vo

_

1kΩ

vin

3V

+

vo

_

1kΩ

vin

Case I: vin > 0

Both diodes are on, and act as short circuits. The equivalent circuit is shown here.

vo = vin

_

+

_

Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.

1kΩ

1kΩ

+

vo

_

+

vo

_

1kΩ

1kΩ

vin

vin

Case II: vin < 0

Both diodes are reverse biased and vo is the sum of the voltage drops across Zener diode and 1kΩ resistor.

3V

3V

_

+

_

Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.

Case II: vin < 0

Both diodes are reverse biased and vo is the sum of the voltage drops across Zener diode and 1kΩ resistor.

1kΩ

+

vo

_

1kΩ

vin

3V

1kΩ

Case IIa: -3V < vin < 0

vo = vin, because the current through Zener diode is zero, all negative voltage drop is across the Zener diode.

+

vo

_

1kΩ

vin

_

+

_

Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.

vo

1kΩ

+

vo

_

1

1kΩ

1

vin

vin

-3V

-3V

-3V

Case IIb: vin < -3V

Excess voltage below -3V is dropped across the two resistors (1kW and 1kW), with

vo = (1/2)*(vin+3)-3= vin/2-1.5 [V].

1

2

(a)

5V

-

+

4V

Ia

-

D

(a)

5V

-

+

4V

Ia

G

-

S

S

G

+

Ib

3V

-

D

+

1V

-

S

G

+

Ic

5V

-

c

D

-

4V

+

D

Id

G

+

1V

+

-

S

3V

-