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
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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,
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,
In the figure shown below,
As the switch (shown in the green) is pressed, the voltage across the diode(ideally)
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
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
In the figure shown below,
When the switch is open, the voltage across the diode (ideally)
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
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
For the initially relaxed circuit shown below, if Ω=1/√LC. Then the current is a function of ___
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
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
When the switch (shown in green) is closed, the average current through the diode in the positive cycle is
When the switch is closed, the steady state current through the diode is
When the switch is open, the current through the diode in the positive cycle is
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
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
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
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
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.
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
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
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
Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle
When a diode is connected in series with an AC source & R load, the conduction time per cycle is
When diode is connected in series to an AC source & RL load, the conduction time for the diode
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,