Power Electronics Multiple Choice Questions on “Diodes Circuits-2”.
1. 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
A. Vs
B. (Vs x s)/RC
C. Vs/s
D. RC/s
Answer: C
Clarification: 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.
2. 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
A. sL + 1/C
B. Cs + 1/sL
C. sL + 1/sC
D. Vs/CLs
Answer: C
Clarification: 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.
3. For the initially relaxed circuit shown below, if Ω=1/√LC. Then the current is a function of ___
A. cos Ωt
B. sin Ωt
C. tan Ωt
D. cos Ωt.sin Ωt
Answer: B
Clarification: 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/Ω2 + s2)
Taking the inverse lapace gives,
I(t) = Vs * √ (L/C. * sin Ωt.
4. 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
A. increases from zero to Vo/R
B. decreases from Vo/R to zero
C. increases from zero to Vo/C
D. decreases from Vo/C to zero
Answer: B
Clarification: 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.
5. 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
A. Vo/C
B. Vo/R
C. Vo
D. Vo/(RC.
Answer: B
Clarification: 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.
6. When the switch (shown in green) is closed, the average current through the diode in the positive cycle is
A. 0
B. Vs/R
C. Vs/(R+RD.
D. none of the mentioned
Answer: A
Clarification: The switch S.C’s the source.
7. When the switch is closed, the steady state current through the diode is
A. Vo/C
B. Vo/R
C. Vo
D. Vo/(RC.
Answer: B
Clarification: I = Vo/R
8. When the switch is open, the current through the diode in the positive cycle is
A. zero
B. Vs/R
C. Vs/(R+RD.
D. none of the mentioned
Answer: C
Clarification: When the switch is open, the diode is forward biased and I = Vs/(Rd + R). Where, Rd is the diode resistance.
9. 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
A. zero value in the positive half cycle and a peak value of 1.414Vs in the negative half cycle
B. sine-wave nature with a peak value 1.414Vs
C. zero value in the negative half cycle and a peak value of 1.414Vs in the positive half cycle
D. sine-wave nature with a peak value Vs