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 Three capacitances, each of C/3 Farads are connected in delta. Their equivalent star value for each capacitance is
  • a)
    3C
  • b)
    C
  • c)
    2/3C
  • d)
    2C
Correct answer is option 'B'. Can you explain this answer?
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Equivalent Star Value of Capacitances Connected in Delta

When three capacitances, each of C/3 Farads, are connected in delta, we need to find their equivalent star value. The equivalent capacitance in star (C_s) can be calculated using the formula:

C_s = C_d/(C_a + C_b + C_c)

where C_d is the delta capacitance and C_a, C_b, and C_c are the individual capacitances connected in delta.

Step-by-Step Solution:

Step 1: Given that each capacitance is C/3 Farads.

Step 2: To find the delta capacitance (C_d), we need to calculate the sum of all three capacitances connected in delta.

C_d = C_a + C_b + C_c

Substituting the given values, we have:

C_d = (C/3) + (C/3) + (C/3)

C_d = C

Step 3: Now, we can substitute the values of C_d, C_a, C_b, and C_c in the formula for the equivalent star capacitance (C_s).

C_s = C_d/(C_a + C_b + C_c)

C_s = C/(C/3 + C/3 + C/3)

C_s = C/(3C/3)

C_s = C/C

C_s = 1

Step 4: The equivalent star value of the three capacitances connected in delta is 1 Farad.

Conclusion:

The correct answer is option 'B' - C. The equivalent star value for each capacitance connected in delta is C.
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it depends on the length of the conductor the capacitance of the line is proportional to the length of the transmission line their effect is negligible on the performance of short having a length less than 80 km and low voltage transmission accidents of the transmission line along with the conductances forms the shunted mittens the conductance and the transmission line is because of the leakage over the surface of the conductor considered a line consisting of two conductors and be each of radius are the distance between the conductors being Des shown in the diagram below minus the potential difference between the conductors and via's work QA charge on conductor QB charge on conductor vvab pencil difference between conductor and the Epsilon minus absolute primitivity QA plus QV = 0 so that QA equals QB - equals DBA equals data equals DB equals our substituting these values and voltage equation we get the capacitance between the conductors is cab is referred to as lying to line capacitance if the two conductors are in VR oppositely charge then the potential difference between them is zero then the potential of each conductor is given by one half bath the capacitance between each conductor and point of zero potential and is capacitive CN is called the capacitance to neut or capacitance to ground capacitance cab is the combination of two equal capacity and VN series thus capacitance to neutral is twice the capacitance between the conductors IE CN equals to Cave the absolute primitivity Epsilon is given by Epsilon equals epsilono Epsilon are where epsilano is the permittivity of the free space and Epsilon or is the relative primitivity of the medium prayer capacitance reactants between one conductor and neutral capacitance of the symmetrical three phase line let a balanced system of voltage be applied to a symmetrical three-phase line shown below the phasor diagram of the three phase line with equilateral spacing is shown below take the voltage of conductor to neutral as a reference phaser the potential difference between conductor and we can be written the similarly potential difference between conductors and sea is on adding equations one and two we get also combining equation three and four from equation 6 and 7 the line to neutral capacitance the capacitance of symmetrical three phase line is same as that of the two wire line Related: Capacitance of Transmission Lines?

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Three capacitances, each of C/3 Farads are connected in delta. Their equivalent star value for each capacitance isa)3Cb)Cc)2/3Cd)2CCorrect answer is option 'B'. Can you explain this answer?
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