Basic Galvanometer MCQ Quiz in मल्याळम - Objective Question with Answer for Basic Galvanometer - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 21, 2025
Latest Basic Galvanometer MCQ Objective Questions
Top Basic Galvanometer MCQ Objective Questions
Basic Galvanometer Question 1:
In a flux meter
Answer (Detailed Solution Below)
Basic Galvanometer Question 1 Detailed Solution
No control torque (negligible) is provided in a flux meter.
Flux Meter
- Flux meter is a special type of ballistic galvanometer having very small control torque and very high electromagnetic damping. Its construction is similar to a moving coil milli-ammeter.
- A coil of the small cross-section is suspended from spring support using a silk thread. The coil moves in the gap of a permanent magnet.
- No control springs are used in a flux meter.
Principle of Operation
- When the flux linking with the search coil is changed an emf is induced in it. Due to this induced emf, a current starts flowing through the flux meter which deflects through an angle depending upon the change in the value of flux linkages.
- The change in the value of the flux is directly proportional to the change in the deflection i.e. \(\phi = \left( {\frac{G}{N}} \right)\theta \).
Therefore, the flux meter has a uniform scale.
Advantages of flux meter:
1. It is a portable instrument.
2. Scale is uniform
3. It can be directly calibrated in weber turns.
4. The deflection of the flux meter is independent of the time taken for the flux to change.
Disadvantages of flux meter:
It is less accurate and sensitive than the ballistic galvanometer.
Basic Galvanometer Question 2:
A supply of 500 Hz is given between terminal a and c and detector is connected between b and d at balance: R2 = 5.8 Ω, R3 = 2200 Ω, R4 = 3000 Ω and C2 = 0.6 μF and r2 = 0.5 Ω. Calculate the dissipating factor of this capacitor.
Answer (Detailed Solution Below)
Basic Galvanometer Question 2 Detailed Solution
At balance
\(\left( {{r_1} + \frac{1}{{J\omega {C_1}}}} \right){R_4} = \left( {{r_2} + {R_2} + \frac{1}{{J\omega {C_2}}}} \right){R_3}\)
Separating the real and imaginary part
\({r_1} = \left( {{r_2} + {R_2}} \right)\frac{{{R_3}}}{{{R_4}}} = \left( {0.5 + 5.8} \right)\frac{{2200}}{{3000}}\)
= 4.62 Ω
And, \({C_1} = {C_2}\frac{{{R_4}}}{{{R_3}}} = \left( {0.6 \times {{10}^{ - 6}}} \right) \times \frac{{3000}}{{2200}}\)
= 0.818 μF
Dissipating factor D1 = tan δ1 = ωC1r1
= 2π × 500 × 0.818 × 10-6 × 4.62
= 11.867 × 10-3
Basic Galvanometer Question 3:
When a current of 1 mA is passed through a ballistic galvanometer, it gives a steady deflection of 45 scale divisions. A capacitor is charged at a pressure of 150 V and discharged through the galvanometer which gives a throw of 300 scale divisions. The complete period of swing is 5 seconds. The capacity of the capacitor is____(in μF)
Answer (Detailed Solution Below) 35 - 36
Basic Galvanometer Question 3 Detailed Solution
Basic Galvanometer Question 4:
When a current of 2 mA is passed through a ballistic galvanometer. It gives a steady deflection of 30 scale divisions A capacitors is charged at a potential of 200 V and discharged through the galvanometer which gives a throw of 300 scale divisions. The complete period of swing is 5 seconds. The capacity of the capacitor is ______________(in μF)
Answer (Detailed Solution Below) 79.5 - 79.7
Basic Galvanometer Question 4 Detailed Solution
\(\begin{array}{l} Q = \frac{I}{\theta }\frac{{{T_o}}}{{2\pi }}{\theta _0}\\ \Rightarrow Q = \frac{{2 \times {{10}^{ - 3}}}}{{30}} \times \frac{5}{{2\pi }} \times \left( {300} \right)\\ \Rightarrow Q = 15.92 \times 10 - 3\\ C = \frac{Q}{V} = \frac{{15.92 \times {{10}^{ - 3}}}}{{200}}\\ \Rightarrow C = 79.58\mu F \end{array}\)
Basic Galvanometer Question 5:
A highly sensitive galvanometer can detect a current as low as 0.1 nA, this galvanometer is used in a s wheat stone bridge as a detector each arm of the bridge as a resistance of 1 KΩ The input voltage applied to the bridge is 20 V. Calculate the smallest change in resistance which can be detected assuming the resistance of the galvanometer is negligible – (in mΩ)
Answer (Detailed Solution Below) 0.02
Basic Galvanometer Question 5 Detailed Solution
\(\begin{array}{l} {V_{TH}} = \frac{{E{\rm{\Delta }}R}}{{4R}}.{R_{th}} = R;{I_g} = \frac{{{V_{TH}}}}{{{R_{th}}}} = \frac{{E{\rm{\Delta }}R}}{{4{R^2}}}\\ \Rightarrow {\rm{\Delta }}R = \frac{{0.1 \times {{10}^{ - 9}} \times 4 \times {{10}^6}}}{{20}} = 0.02\ m{\rm{\Omega }} \end{array}\)