Corona MCQ Quiz in தமிழ் - Objective Question with Answer for Corona - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 17, 2025
Latest Corona MCQ Objective Questions
Top Corona MCQ Objective Questions
Corona Question 1:
The good effect of corona on overhead lines is to -
Answer (Detailed Solution Below)
Corona Question 1 Detailed Solution
Good Effect of Corona on Overhead Lines:
Corona discharge occurs when the electric field intensity around the conductor exceeds the dielectric strength of the surrounding air, leading to ionization of the air and formation of a visible glow or corona. Although corona discharge is generally undesirable because it results in power loss, audible noise, and radio interference, it has a beneficial effect in certain situations.
The key beneficial effect of corona on overhead lines is:
- Reduction of the Steepness of Surge Fronts: During a lightning strike or switching operation, high-voltage surges travel along the transmission lines. These surges can have steep wavefronts which pose a significant threat to insulation and equipment. The presence of corona discharge around the conductors can help to reduce the steepness of these surge fronts. The ionized air around the conductor acts as a cushion, dissipating energy and smoothing out the sharp edges of the surges. This results in less severe surges, which are less likely to cause damage to the insulation and other components of the power system.
Therefore, the correct answer is option 4.
Corona Question 2:
At 50 Hz operating frequency a HVAC transmission has a corona loss of 1.5 kW/ph/km. What is the corona loss of the same line operating at 60 Hz._______ (in kW/ph/kM)
Answer (Detailed Solution Below) 1.7
Corona Question 2 Detailed Solution
Corona loss, Ploss α (f + 25)
\(\begin{array}{l} \Rightarrow \frac{{1.5}}{{{{\left( {{P_{loss}}} \right)}_{{{60}_{Hz}}}}}} = \frac{{\left( {50 + 25} \right)}}{{\left( {60 + 25} \right)}}\\ \Rightarrow {\left( {{P_{loss}}} \right)_{{{60}_{Hz}}}} = \frac{{\left( {85} \right)\left( {1.5} \right)}}{{75}} = 1.7\ kW/ph/kM \end{array}\)
Corona Question 3:
A 3 phase line operating at 150 kV which has conductor of 1.5 cm diameter arranged in a 4 meter delta. Assume air density factor of 1.2 and the dielectric strength of air to be \(21\frac{{kV}}{{cm}}\). Calculate the corona loss ____ \(\left( {in\ \frac{{kW}}{{kM}}/Phase} \right)\).
Answer (Detailed Solution Below) 0
Corona Question 3 Detailed Solution
\(\begin{array}{l} V = 21m\delta r\ln \frac{d}{r}\\ = 21 \times 1.2 \times 1 \times 0.75\ \ln \left( {\frac{{400}}{{0.75}}} \right) \end{array}\)
= 118.68 kV
Line to line voltage is = 205.55 kV
Since the operating voltage is 150 kV, the corona loss will be absent.
⇒ Corona loss = 0.