Example Solved Numerical Problems

EXAMPLE 9. 10

What is the output Y in the following circuit, when all the three inputs A, B, and C are first 0 and then 1?

Solution

EXAMPLE 9. 11

In the combination of the following gates, write the Boolean equation for output Y in terms of inputs A and B.

Solution

The output at the 1st AND gate: A

The output at the 2nd AND gate: B 

The output at the OR gate: Y = A.  + . B

De Morgan’s Theorem - Numerical Problems Questions with Answers, Solution

EXAMPLE: 9 . 12

Simplify the Boolean identity

AC + ABC = AC

Solution

Step 1: AC (1 + B) = AC.1 [OR law-2]

Step 2: AC . 1 = AC [AND law – 2]

Therefore, AC + ABC = AC

Circuit Description

Thus the given statement is proved.

Space Wave Propagation - Numerical Problems Questions with Answers, Solution

EXAMPLE 10.1

A transmitting antenna has a height of 40 m and the height of the receiving antenna is 30 m. What is the maximum distance between them for line-of-sight communication? The radius of the earth is 6.4×106 m.

Solution:

The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage.

d = d1 +d2

= √(2Rh­) + √(2Rh2)

= √2√R (√h1 + âˆšh2)

= √[2 × 6.4 ×106] × ( √40 + √30 )

=16 ×102 âˆš5 ×(6.32 + 5.48)

=42217m= 42.217 km

Numerical Problems

1. The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance R1 [Ans: 2.5 A]

Solution:

Barrier potential for ideal diode is zero. The diode D1 is reverse biased, so it will block the current and diode D2 is forward biased, so it will pass the current.

The given circuit becomes

Effective resistance Reff = R1 + R3 = 4Ω

Current through R1 = V/Reff = 10/4 = 2.5A

2. Four silicon diodes and a 10 Ω resistor are connected as shown in figure below. Each diode has a resistance of 1Ω. Find the current flows through the 18Ω resistor. [Ans: 0.13 A]

Solution:

In the given circuit D2 & D3 are in forward bias so they conduct current while D1 &D4 are in reverse bias so they do not conduct current. So the equivalents circuit will be

the effective resistance is Reff

= 1Ω + 10Ω + 1Ω = 12Ω

Here silicon diodes are used

∴ Barrier potential for Si is 0.7 V

Net potential (Vnet) = 3 – 0.7 - 0.7

Vnet = 1.6 V

Current (I) = Vnet / R eff = 1.6 / 12 = 0.133A

3. Assuming VCEsat = 0.2 V and β = 50, find the minimum base current (IB) required to drive the transistor given in the figure to saturation. [Ans: 56 µA]

Solution;

VCE = 0.2V

Rc = 1kΩ

β =50

Vcc = 3V

I B = ?

Ic = [ VCC â€“ VCE ] / RC

= (3-0.2) / 103

= 2.8 × 10-3 A

= 2.8 mA

β= IC/IB

∴ IB = IC / β = 2.8 × 10-3 / 50 = 0.056× 10-3

= 56 × 10-6 A

IB=56μA

4. A transistor having α =0.99 and VBE = 0.7V, is given in the circuit. Find the value of the collector current. 

[Ans: 5.33 mA]

Given data:

α =0.99

VBE = 0.7V

IC=?

Solution:

Tranistor is in saturation region.

∴ Ic = Ic(sat)

IC and IB are independent

VCE(sat) = 0.2 V, VBE(sat) = 0.8 V for silicon transistor (ie, standard value)

Apply KVR across B-E loop:

V1k + V10k + VBE sat + V1k = VCC

∴ 1(IC + IB) + 10 IB + 0.8 + 1 (IC + IB) = 12

2 IC + 12 IB = 11.2         ……………(1)

Apply KVR across C-E loop:

V1k + V1k + VCE sat + V1k = VCC

l(IC + IB) + 1 IC + 0.2 + I(IC + IB) = 12

3IC + 2IB = 11.8    ……………….(2)

Solve the equation (1) and (2)

IB = 0.3125 mA

IC = 3.725 mA ≈ 3.73 mA

5. In the circuit shown in the figure, the BJT has a current gain (β) of 50. For an emitter – base voltage VEB = 600 mV, calculate the emitter – collector voltage VEC (in volts). [Ans: 2 V]

Given data:

 Î² = 50

VE=3V

VEB=60 mV

RB=60KΩ

RE=500Ω

Solution:

VB=VE−VEB

VB = 3−0.6 = 2.4V

IB = VB/RB = 2.4 / 60×103 = 40μA

IC = βIB = 50×40μA = 2×10-3 A = 2mA

VC = RCIC = 500×2×10-3 = 1V

VEC = VE-VC = 3-1 = 2V

VEC = 2V

6. Determine the current flowing through 3Ω and 4Ω resistors of the circuit given below. Assume that diodes D1 and D2 are ideal diodes.

Solution:

The diode D2 is in reverse biased. So does not conduct current.

∴ current through 3Ω is = 0

The diode D1 is in forward biased, and it is an ideal diode. So the given circuit becomes as

The current through 4Ω is

∴ I = V/R = 12/6 =2A

7. Prove the following Boolean expressions using the laws and theorems of Boolean algebra.

i) (A + B) (A + ) = A

ii) A  ( +B) = AB

iii) (A + B) (A + C) = A + BC

Solution:

8. Verify the given Boolean equation A +  á¾¹B = A + B using truth table.

Solution:

Hence, verified

9. In the given figure of a voltage regulator, a Zener diode of breakdown voltage 15V is employed. Determine the current through the load resistance, the total current and the current through the diode. Use diode approximation.

Solution:

Voltage across RL(VO) = Vz = 15V

 Voltage across RS(VRS) = 25 -15 = 10V

current through RL is

IL = V0/RL = 15 / 3×103 = 5×10-3 A

IL = 5 mA

Current through RS is

I = VRS/RS = 10/500 = 20×10-3A

I = 20mA

Current through Zener diode is

IZ=I-IL

= (20-5) Ã— 10-3

Iz = 15mA

10. Write down Boolean equation for the output Y of the given circuit and give its truth table.

Solution: