## Electronics Quiz-7

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. Two identical p-n junction may be connected in series with a battery in three ways as shown in the adjoining figure. The potential drop across the p-n junctions are equal in

•  Circuit 1 and circuit 2
•  Circuit 2 and circuit 3
•  Circuit 3 and circuit 1
•  Circuit 1 only
Solution
(b) In circuit 1, N is connected with N, which is not a series combination of p-n junction. In circuit 2, each p-n junction is forward biased, hence same current flows, giving same potential difference across p-n junction. In circuit 2, each p-n junction is reverse biased, same leakage current will flow, giving equal potential difference across each p-n junction diode.

Q2.While a collector to emitter voltage is constant in a transistor, the collector current changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio hfe is
•  82
•  83
•  8.2
•  8.3
Solution
(a) hfe=((Î”ic)/(Î”ib))(Vce=8.2/(8.3-8.2)=82

Q3.  The mobility of free electron is greater than that of free holes because
•  The carry negative charge
•  They are light
•  They mutually collide less
•  They require low energy to continue their motion
Solution
(D)

Q4. Any digital circuit can be realised by repetitive use of only
•  NOT
•  OR
•  AND
•  NOR
Solution
(d) NOR and NAND gates are universal gates. Any digital circuit can be realised by repetitive use of these (NOR and NAND) gates.

Q5.In p-n junction, the barrier potential offers resistance to
•  Free electrons in n-region and holes in p–region
•  Free electrons in p-region and holes in n-region
•  Only free electrons in n-region
•  Only holes in p-region
Solution
(a) In p-n junction, the barrier potential offers resistance to free electrons in n-region and holes in p-region.

Q6. Would there be any advantage to adding n-type or p-type impurities to copper
•  YES
•  No
• May be
•  Information is Insufficient
Solution
(b) Pure Cu is already an excellent conductor, since it has a partially filled conduction band, furthermore, Cu forms a metallic crystal as opposed to the covalent crystals of silicon or germanium, so the covalent crystals of silicon or germanium, so the scheme of using an impurity to donate or accept an electron does not work for copper. In fact adding impurities to copper decreases the conductivity because an impurity tends to scatter electrons, impeding the flow of current

Q7.The coordination number of hexagonal close packing (hcp) is
•  6
•  8
•  12
•  16
Solution
(c) Hexagonal close packing is 12 for face centred crystal

Q8.A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.

•

•

•

•

Solution
(C)

Q9.The frequency response curve of RC coupled amplifier is shown in figure. The band width of the amplifier will be

•  f3-f2
•  f4-f1
•  (f4-f2)/2
•  f3-f1
Solution
(c) (b) The band width is defined as the frequency band in which the amplifier gain remains above 1/√2=0.707 of the mid frequency gain (A_max). The low frequency f_1 at which the gain fall to 1/√2,i.e.,0.707 times it’s mid frequency value is called lower cut off frequency and the high frequency f_4 at which the gain falls to 1/√2,i.e.,0.707 times of it’s mid frequency is known as higher cut off frequency so band width =f4-f1

Q10. The typical ionisation energy of a donor in silicon is
•  10.0eV
•  1eV
•  0.1eV
• 0.001eV
Solution
(c) When donor impurity (+5 valence) is added to a pure silicon (+4 valence), the +5 valence donor atom sits in the place of +4 valence silicon atom. So it has a net additional +1 electronic charge. The four valence electrons form covalent bond and get fixed in the lattice. The fifth electron (with net -1 electronic charge) can be approximated to revolve around +1 additional charge. The situation is like the hydrogen atom for which energy is given by E=-13.6/n2eV. For the case of hydrogen, the permittivity was taken as Îµ0. However, if the medium has a permittivity Îµ_r, relative to Îµ_0, then E=-13.6/(Îµ_r2 n2 ) eV For Si,Îµr=12 and for n=1,E≃0.1 eV

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