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This mock test of Test: Diode Circuit for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
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QUESTION: 1

An ideal diode has _________ &__________

Solution:

An ideal diode has no losses and negligible reverse recovery time.

QUESTION: 2

A diode circuit is so arranged that when the switch is open it’s KVL gives

Ri+ 1/C ∫i dt = 0

When the switch is closed,

Ri+ 1/C ∫i dt = Vs

Vs is the dc supply voltage.

The diode is so connected that it is forward biased when switch is closed

The circuit is mostly likely be a

Solution:

Examine the equation, the same current flows through R & C. Also when switch is open, the equation R.H.S is 0. Hence, all the elements are in series.

QUESTION: 3

A circuit is so formed such that the source-R-C-diode-switch are in series. Consider the initial voltage across the C to be zero. The diode is so connected that it is forward biased when the switch is closed. When the switch is closed,

Solution:

Instant switch is closed, the current is maximum Vs/R than starts to reduce, whereas voltage starts to increase from 0 to Vs (Capacitor Charging).

QUESTION: 4

The time constant of a series RC circuit (τ) is given by

Solution:

For a series RC circuit, τ = RC.

QUESTION: 5

A circuit is so formed such that source-R-L-diode-switch are all in series. Consider the initial current in L to be zero. The diode is so connected that it is forward biased when switch is closed.

When the switch is closed,

Solution:

Instant switch is closed, the current is minimum zero than starts to increase till it reaches a constant value Vs/R, whereas voltage starts to reduce from Vs to 0 (Inductor Charging).

QUESTION: 6

In the figure shown below,

As the switch (shown in the green) is pressed, the voltage across the diode(ideally)

Solution:

As the switch is pressed, current starts to flow & the whole supply voltage (ideally) appears across the load R & voltage across the diode is zero.

QUESTION: 7

For a diode circuit the voltage across the capacitor is given by

Vc(t)= Vs(1-e^{(-t/RC)})

Then the initial rate of change of capacitor voltage is given by

Solution:

Find d(Vc)/dt and put t = 0.

QUESTION: 8

In the circuit show below,

The initial current through the inductor is zero. When the switch (shown in green) is closed, then the current through the inductor

Solution:

Current increases from zero to maximum value gradually due to the L nature. The KVL when switch is closed gives,

Ri + Ldi/dt = Vs

Solve for i(t). Maximum value comes out to be Vs/R.

QUESTION: 9

In the figure shown below,

When the switch is open, the voltage across the diode (ideally)

Solution:

When the switch is open, the diode experiences all the supply voltage.

QUESTION: 10

The time constant of a series RL circuit (τ) is given by

Solution:

The time constant τ for a series RL circuit is R/L.

QUESTION: 11

For the initially relaxed circuit shown below, KVL with switch in the closed position gives a certain equation. The Laplace of this equation will have the right hand side (RHS) as

Solution:

When switch is closed,

Ldi/dt + 1/C ∫ idt = Vs

Laplace of the above gives,

L[sI(s)] + 1/C [I(s)/s] = Vs/s.

QUESTION: 12

For the initially relaxed circuit shown below, the Laplace transform of the KVL when the switch is closed is

I(s) [ X ] = Vs/s

The value of X is

Solution:

When switch is closed,

Ldi/dt + 1/C ∫ idt = Vs

Laplace of the above gives,

L[sI(s)] + 1/C [I(s)/s] = Vs/s.

QUESTION: 13

For the initially relaxed circuit shown below, if Ω=1/√LC. Then the current is a function of ___

Solution:

When switch is closed,

Ldi/dt + 1/C ∫ idt = Vs

Laplace of the above gives,

L[sI(s)] + 1/C [I(s)/s] = Vs/s

I(s) = Vs/(LΩ) * (Ω/Ω^{2} + s^{2})

Taking the inverse lapace gives,

I(t) = Vs * √ (L/C) * sin Ωt.

QUESTION: 14

For the circuit shown below, the capacitor is initially charged to a voltage of Vo with the upper plate positive. After the switch (shown in green) is closed, the current through the load

Solution:

The capacitor acts as a source. At instant switch is closed the current is maximum and than discharges till zero value through the load R.

QUESTION: 15

For the circuit shown below, the capacitor is initially charged to a voltage of Vo with the upper plate positive. Switch is closed at t=0. The peak value of the current through the diode is

Solution:

When switch is closed, the equation is

Ri + 1/C ∫idt = 0 (Voltage across capacitor cannot change instantaneously)

Solution of the above equation gives, Vo/R at t= 0.

QUESTION: 16

When the switch (shown in green) is closed, the average current through the diode in the positive cycle is

Solution:

The switch S.C’s the source.

QUESTION: 17

When the switch is closed, the steady state current through the diode is

Solution:

I = Vo/R

QUESTION: 18

When the switch is open, the current through the diode in the positive cycle is

Solution:

When the switch is open, the diode is forward biased and I = Vs/(Rd + R). Where, Rd is the diode resistance.

QUESTION: 19

For the circuit shown in the figure below, consider the diode as an ideal diode & R.M.S value of source voltage as Vs.

The output voltage waveform at R is most likely to have

Solution:

The diode S.C’s the load in the positive half cycle.

QUESTION: 20

For the circuit shown in the figure below, V8 is AC voltage source with peak value Vm. The waveform of the load voltage at the resistor is

Solution:

Diode is reversed biased in the positive half cycle. In the negative half cycle, apply KVL to get the value of peak voltage at the load.

QUESTION: 21

For the circuit shown in the figure below, consider the diode as an ideal diode & rms value of source voltage as Vs.

The output voltage waveform at R will have

Solution:

The diode short circuits the load in the negative half cycle. The peak value in the positive half is 1.414 x Vs.

QUESTION: 22

For the circuit shown in the figure below, Vs is the ac voltage source with peak value Vm. The waveform of the load voltage at the resistor will have

Solution:

Diode is reversed biased in the positive half cycle. In the negative half cycle, apply KVL to get the value of peak voltage at the load.

QUESTION: 23

For the circuit shown below,

Vs=230V, Voltage drop across the diode (Vd) = 2V

The peak value of voltage at R in the positive and the negative half cycles are ___ & ___ respectively.

Solution:

Load is S.C in the positive half cycle hence, voltage is zero. In negative half the peak value is (230 x 1.414) – 2(drop across the diode) = 323 Volts.

QUESTION: 24

For the circuit shown in the figure below, V9 is the AC voltage source with peak value Vm. The waveform of the load voltage at the resistor has a

Solution:

Diode is reversed biased in the negative half cycle. In the positive half cycle, apply KVL to get the value of peak voltage at the load. Peak value = Vm + V1, as V1 is aiding the positive anode of the diode.

QUESTION: 25

The circuit shown below has the following parameters:

V1 = 8 Volts

V2 = 6 Volts

Vs = 10V/√2 (rms)

Voltage drop across D1 & D2 = 0.7 Volts

At the load (R), the peak value in the positive half cycle will be

Solution:

In the positive half, only D1 is active. Hence use KVL, Vo = 8 + 0.7 Volts.

QUESTION: 26

The circuit shown below has the following parameters:

Voltage drop across D1 & D2 = 0.7 Volts

V1 = 8 Volts

V2 = 6 Volts

Vs = 10V/√2 (RMS)

At the load (R), the peak value in the negative half cycle is

Solution:

In the negative half, only D2 is active. Hence use KVL, Vo = 6 + 0.7 Volts.

QUESTION: 27

Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle

Solution:

During the negative half diode is not conducting hence Vr (voltage across the resistor) = 0. Vo = Vdc.

QUESTION: 28

When a diode is connected in series with an AC source & R load, the conduction time per cycle is

Solution:

The diode is forward biased for half cycle (180 degrees) and reversed biased in the another 180 degrees. Hence, per cycle it only conducts for 180 degrees or π radians.

QUESTION: 29

When diode is connected in series to an AC source & RL load, the conduction time for the diode

Solution:

For an R – L load, the inductively load can make the diode to force conducted hence, the conduction time can be greater than 180 degrees.

QUESTION: 30

For the circuit shown below,

Vdc = 50 V

Cut-in voltage for D1 = 0.2 V

Cut-in voltage for D2 = 0.6 V

R = 5KΩ

Current through D1 & D2 would be,

Solution:

As the cut-in voltage for D1 is lesser than D2, D1 would start conducting first & S.C D2. Hence, only D1 conducts with I = 50/5000 A.

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