Sample Exam 3

ERP = Effective Radiated Power, apply the formula, Power gain (dB) = 10 Log( P₁ / P₂), Power gain = 10 dB, P₁ = 100 W and find P₂ , therefore 10 = 10 Log (100 / P₂), therefore 1 = Log (100 / P₂), Log 10 = Log (100 / P₂), 10 = 100 / P₂ therefore P₂ = 10

If the signal level is high and noise level low, that means S/N is high and you can have a good reception, If S/N is low the reception will be poor.

First you must know the resistor color code:- 0 = black, 1 = brown, 2 = red, 3 = orange, 4 = yellow, 5 = green, 6 = blue, 7 = violet, 8 = gray, 9 = white; In this question, 1st color band is red represent 2 and the 2nd green represents 5 and 3rd brown represents 1 and it’s the multiplyer. It is equal to 10¹ . So the value is 250. If it is Orange, blue, green , the value is 3600000 Ω or 3.6 MΩ.

LF (low frequency or long wave) = 30 to 300 kHz. ; MF (medium frequency or medium wave) = 300 to 3000 kHz. ; HF (high frequency or short wave) = 3 to 30 MHz. ; VHF (very high frequency) = 30 to 300 MHz. ; UHF (ultra high frequency) = 300 to 3000 MHz. ; SHF (super high frequency) = 3 to 30 GHz.; EHF ( extra high frequency or micro wave) = 30 to 300 GHz.

Class A is Full cycle, class B is half cycle, class C is Quater or less than half cycle. In between there are some little bit different classes like, class AB, AB1 ect.

Reflection coefficient = sqr(Ref.pwr / Forw. Pwr) = Ref. Voltage/Forw. Voltage, ;Perfectly matched antenna has no reflected power, so Reflected coefficent = 0, Total power is rflect in perfectly mismatch antenna , so Reflected coefficent = 1, Always reflected coefficient having a value between 0 and 1. It will not take a value of -1 or infinity.

As same as Resistances. Parallel two inductors having 4L (lower side) is equivalent to 2L, This 2L and other 2L are parallel. It is equivalent to 1L. This 1L is in series with other 3L and L. It is equivalent to 5L.

If you connect a dc meter with correct polarity, indicator is moving correct way ( clockwise). If it is on wrong polarity indicator slightly move wrong way (anti clockwise) and getting stuck. Power or power factor or RMS value canot measure with a dc voltmeter.

three resistore of 3R are connected in parallel and equivalent to R, two resistors of 2R are also connected in parallel and equivalent to R. these two resistors of R are in series having a value of 2R.

When 100W of SSB signal and 400 W AM signal are comming from the same station can receive approximately with equal strength.

BFO use to demodulate SSB and CW signals.

Use the formula 1/R = 1 / R₁ + 1 / R₂ + 1 / R₃, ; 1 / R = 1/12+1/15+1/20 = 5/60+ 4/60+ 3/60 = 12/60 = 1/5, therefore R = 5

Apply C = C₁ + C₂ + C₃ for parallel capacitors, therefore C = 8 + 4 + 2 = 14

Series combination of 10 μH and 20 μH is equivalent to 30 μH. Same way other two equivalent to 70 μH. Parrallel combination of 30 μH and 70 μH is equivalent to 30 x 70 / ( 30+70) = 2100/100 = 21 μH.

Apply the formula, f = 1 / {2π √(LC)} therefore f = 1 / {2π √(100x10ˉ⁶ x 100x10-12)} = 1 / 2π x 10-7 = (10/2π) x 10-6 Hz = (10/2π) MHz = 1.6 MHz

27Ω and 33 Ω are in series. It is equivalent to 60 Ω. The current flow through this 60Ω is equal to the current flow through the 27Ω resistor. Apply V = I R to that 60 Ω, V=6, R=60, therefore 6 = I x 60 , therefore I = 6/60 = 0.1 A = 100 mA

An AC current passing through the speech coil. Therefore the power is equal to VI or I² Z or V² / Z , V = 4Volt, Z = 4Ω,. Therefore Power = 4² / 4 = 4 Watt

Ohm is the unit of resistance, Henry is the unit of Inductance, Farad is the unit of capacitance, Hertz is the unit of frequency.

Peak value = √2 x RMS value or RMS value = (1/√2) x Peak value.

Apply the formula R = ρ l / A, for a conducter. “ ρ “ is the resistivity of the material, “ l “ is the length and “ A “ is the aera of the cross section. According to the formula R is doubled, if length doubled or A become half. “ ρ “ is constant for a particular material.

Two series resistors of 4Ω equivelent to 8Ω. This one and other 8 Ω resistor connected in parallel and equivalent to 4Ω. This 4Ω combined in series with two resistors of 2Ω. It is equivalent to 8Ω.

First calculate the equivalent resistance between A and B. Two resistors of 2 Ω (Right hand and upper side) are equivalent to 4Ω. This one is parallel with othe 4Ω in the circuit. This parallel combination equivalent to 2Ω and it is connected in parellel with other 2Ω (middle one) and equivalent to 1Ω. This one in series with remain 2Ω and the final value is 3Ω. Then apply Ohm’s law to this 3Ω resistor. V = 3 Volt, R = 3Ω. Therefore I = V / R = 3/3 = 1

There are no semiconductors like PPN, NNP, PNN, NPP, PPP, NNN.

Two capacitors equal to C are in parallel and they are equivalent to 2C. This 2C connected in series with other two capacitore having a value of 2C and it is equivalent to 2C/3.

Two inductors having 2H are in parallel. They are equivalent to H. This is in series with other H and equivalenet to 2H.

apply the formula f (MHz) x wave length (m) = 300 , (speed of light = f λ). Therefore 100 λ =300, λ = 300/100 = 3, λ/2 = 3/2 = 1.5 m

Apply P = I² R , P = 0.25 W, R = 400Ω. Therefore 0.25 = I² x 400, therefore I² = 0.25 / 400 = 25 / 40000 And I = 5/200 A = 25 mA

Apply the formula power gain (dB) = 10 Log(P₁ / P₂) Therefore 3 = 10 Log (P₁ / P₂) , therefore 0.3 = Log (P₁ / P₂) , Log 2 = 0.3010, approximatly Log 2 = 0. 3 , therefore Log 2 = Log(P₁ / P₂) , Therefore 2 = (P₁ / P₂) , double the power.

Apply P = VI therefore P = 24 x 2.5 = 60

Naturally it is 300Ω, Simple dipole is 75Ω

apply the formula f (MHz) x wave length (m) = 300 , (speed of light = f λ). Therefore 60λ =300 , λ = 300/60 = 5 m.

Apply the formula X = 2 π f L, If f increases X also increase.

Power factor lying between 0 and 1. For a capacitor PF = 0

The effective value three 3Ω resistors is 1Ω, because they are in parallel. This 1Ω is connected in series with the other 1Ω resistor in the circuit and getting the equivalent value of 2Ω. Therefore the power dissipation is = V² / R = 2² / 2 = 2 W

apply the formula R = ρ l / A, ρ is the resistivity of the material, it is a constant. If the conductivity decrease, Resistance will be increased. When you consider this formula, R increased, if “ l “ increase or “A” decrease. “A” decrease mean reducing the diameter.

Balance modulater giving a DSB signal. When it is passing through the side band filter , output will be USB or LSB accordingly.

Use the formula E = ½ C V², C = 1 F, V = 2 volts , therefore E = ½ x 1 x 2² = 2 Jule

Use the formula Z² = R² + X² , R = 3 Ω, X = 4 Ω, therefore Z² = 3² + 4² = 9 + 16 = 25 , therefore Z = 5 Ω