ELECTRONICS-QUIZ 11
Dear Readers,
As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. If a full wave rectifier circuit is operating from 50Hz mains, the fundamental frequency in the ripple will be
Solution
(b) For full wave rectifier, ripple frequency = 2× input frequency = 2 × 50= 100 Hz
(b) For full wave rectifier, ripple frequency = 2× input frequency = 2 × 50= 100 Hz
Q2.In a cubic unit cell of bcc structure, the lattice points (ie, number of atoms) are
Solution
(a) Number of lattice points in a crystal structure will be n=Nc/8+Nf/2+Ni/1 In bcc crystal, Nc=8,Nf=0 and Ni=1 n=8/8+0/1+1/1=2
(a) Number of lattice points in a crystal structure will be n=Nc/8+Nf/2+Ni/1 In bcc crystal, Nc=8,Nf=0 and Ni=1 n=8/8+0/1+1/1=2
Q3. The output in the circuit of figure is taken across a capacitor. It is as shown in figure
Solution
(c) As RC time constant of the capacitor is quite large (Ï„=RC=10×103×10×10-6=0.1 s), it will not discharge appreciably. Hence voltage remains nearly constant
(c) As RC time constant of the capacitor is quite large (Ï„=RC=10×103×10×10-6=0.1 s), it will not discharge appreciably. Hence voltage remains nearly constant
Q4. Two identical capacitors A and B are charged to the same potential V and are connected in two circuits at t=0, as shown in figure. The charge on the capacitors at time t=CR are respectively
Solution
(b) Time t=CR is known as time constant. It is time in which charge on the capacitor decreases to 1/e times of it’s initial charge (steady state charge). In figure (i) PN junction diode is in forward bias, so current will flow the circuit i.e., charge on the capacitor decrease and in time t it becomes Q=1/e (Q_o ); where Q_o=CV⇒Q=CV/e In figure (ii) P-N junction diode is in reverse bias, so no current will flow through the circuit hence change on capacitor will not decay and it remains same, i.e.,CV after time t
(b) Time t=CR is known as time constant. It is time in which charge on the capacitor decreases to 1/e times of it’s initial charge (steady state charge). In figure (i) PN junction diode is in forward bias, so current will flow the circuit i.e., charge on the capacitor decrease and in time t it becomes Q=1/e (Q_o ); where Q_o=CV⇒Q=CV/e In figure (ii) P-N junction diode is in reverse bias, so no current will flow through the circuit hence change on capacitor will not decay and it remains same, i.e.,CV after time t
Q5.If De , Db and Dc are the doping levels of emitter, base and collector respectively of a transistor, then
Solution
(d) In transistor emitter is heavily doped and base is lightly doped. So, De > Dc > Db
(d) In transistor emitter is heavily doped and base is lightly doped. So, De > Dc > Db
Q6. The saturation current in a diode valve is governed by
Solution
(c) For a triode, the relation is μ=rp×gm μ= amplification factor rp= plate resistance gm= mutual conductance
(c) For a triode, the relation is μ=rp×gm μ= amplification factor rp= plate resistance gm= mutual conductance
Q7.Which of the following does not vary with plate or grid voltages
Solution
(d)Each of them varies
(d)Each of them varies
Q8.A zener diode has a contact potential of 1 V in the absence of biasing. It undergoes Zener breakdown for an electric field of 106 V-m^(-1) at the depletion region of p-n junction. If the width of the depletion region is 2.5 μm, what should be the reverse biased potential for the Zener breakdown to occur?
Solution
(b) Reverse biased potential for the zener breakdown Vr=Ed=106×2.5×10-6=2.5 volt
(b) Reverse biased potential for the zener breakdown Vr=Ed=106×2.5×10-6=2.5 volt
Q9.In order to prepare a p-type semiconductor, pure silicon can be doped with
Solution
(b) When an impurity atom with 3 valence electrons (as aluminium) is introduced in a pure silicon crystal, all the three of its valence electrons form covalent bonds with one each valence electrons of the nearest
silicon atom while the valence electron of the fourth nearest silicon atom is not able to form the bond, leading to formation of hole or p-type semiconductor. While phosphorus being a pentavalent impurity leads to formation of n-type semiconductor. class="separator" style="clear: both; text-align: center;">
(b) When an impurity atom with 3 valence electrons (as aluminium) is introduced in a pure silicon crystal, all the three of its valence electrons form covalent bonds with one each valence electrons of the nearest
silicon atom while the valence electron of the fourth nearest silicon atom is not able to form the bond, leading to formation of hole or p-type semiconductor. While phosphorus being a pentavalent impurity leads to formation of n-type semiconductor. class="separator" style="clear: both; text-align: center;">
Q10. The output wave form of full-wave rectifier is
Solution
(c)
(c)
