A sinusoidal voltage of peak value 200 volts is connected to a diode and resistor R in the circuit shown so that half-wave rectification occurs. If the forward resistance of the diode is negligible compared to R the r.m.s voltage (in volt) across R is approximately

(1) 200                           

(2) 100

(3) 2002                         

(4) 280

Subtopic:  PN junction |
From NCERT
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The junction diode in the following circuit requires a minimum current of 1 mA to be above the knee point (0.7 V) of its I-V characteristic curve. The voltage across the diode is independent of current above the knee point. If VB = 5 V, then the maximum value of R so that the voltage is above the knee point, will be

(1) 4.3 kΩ                

(2) 860 kΩ

(3) 4.3 Ω                  

(4) 860 Ω

Subtopic:  PN junction |
 80%
From NCERT
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In the circuit given below, V(t) is the sinusoidal voltage source, voltage drop VAB(t) across the resistance R is

(1) Is half wave rectified

(2) Is full wave rectified

(3) Has the same peak value in the positive and negative half cycles

(4) Has different peak values during positive and negative half cycle

Subtopic:  Applications of PN junction |
From NCERT
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The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10 V. The dc component of the output voltage is 
(1)10/2  V                                   

(2) 10/π V

(3) 10 V                                           

(4) 20/π V

Subtopic:  Rectifier |
 56%
From NCERT
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A transistor is used as an amplifier in CB mode with a load resistance of 5 kΩ. The current gain of the amplifier is 0.98 and the input resistance is 70 Ω, the voltage gain and power gain respectively are :
(1) 70, 68.6                         

(2) 80, 75.6

(3) 60, 66.6                         

(4) 90, 96.6

Subtopic:  Applications of Transistor (OLD NCERT) |
 79%
From NCERT
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In the following circuits PN-junction diodes D1D2 and D3 are ideal for the following potential of A and B, the correct increasing order of resistance between A and B will be

(i) – 10 V, – 5V           (ii) – 5V, – 10 V
(iii) – 4V, – 12V

(a) (i) < (ii) < (iii)                       (b) (iii) < (ii) < (i)
(c) (ii) = (iii) < (i)                       (d) (i) = (iii) < (ii)

Subtopic:  Applications of PN junction |
 63%
From NCERT
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The circuit shown in the following figure contains two diode D1 and D2 each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100 ohm resistance (in amperes) is

1.  Zero 
2.  0.02
3.  0.03 
4.  0.036

Subtopic:  PN junction |
 80%
From NCERT
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The contribution in the total current flowing through a semiconductor due to electrons and holes are 34 and 14 respectively. If the drift velocity of electrons is 52 times that of holes at this temperature, then the ratio of concentration of electrons and holes is
1. 6 : 5                 

2. 5 : 6

3. 3 : 2                 

4. 2 : 3

Subtopic:  Types of Semiconductors |
 67%
From NCERT
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Ge and Si diodes conduct at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the valve of V0 changes by 

(1) 0.2 V                   

(2) 0.4 V

(3) 0.6 V                   

(4) 0.8 V

Subtopic:  PN junction |
 69%
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In the following circuit, find I1 and I2.

(a) 0, 0
(b) 5 mA, 5 mA
(c) 5 mA, 0
(d) 0, 5 mA

Subtopic:  Applications of PN junction |
 53%
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