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The "Ohm's law" for the magnetic circuit States that the net magneto-motive force (mmf) equals the product of the core reluctance and the core flux.

(a) True

(b) False

Short Answer

Expert verified

Therefore, the correction option is (a): True

Step by step solution

01

State the ohm’s law for magnetic circuit.

The ohms says that the MMF in the circuit is directly proportional to the magnetic flus in the circuit.

Fαϕ

F=ϕR

Here, Fis the MMF,ϕis the flus and is the proportionality constant reluctance.

The MMF in the current is defined as the product of the number of the turns and current flowing through it.

F=Nl

02

Find the net magneto motive force in the circuit.

The ohm’s law equation for the magnetic circuit is given by,

N1l1-N2l2=Rϕ

From the above equation, the net MMF is directly to product of reluctance and core flux.

Therefore, the correction option is (a): True

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Most popular questions from this chapter

A three-phase transformer is rated 1000MVA,220Y/22ΔkV. The Y-equivalent short-circuit impedance, considered equal to the leakage reactance, measured on the low-voltage side is0.1Ω.Compute the per-unit reactance of the transformer. In a system in which the base on the high-voltage side of the transformer is100MVA,230kV, what value of the per-unit reactance should be used to represent this transformer?

The ratings of a three-phase three-winding transformer are

Primary (1) : Y connected, 66kV, 15MVA

Secondary (2) : Y connected, 13.2kV, 10MVA

Tertiary (3): A connected, 2.3kV, 5MVA

Neglecting winding resistances and exciting current, the per-unit leakage reactances are

X12 = 0.08on a 15MVA, 66kV base

X13 = 0.10on a 10 MVA, 13.2kV base

X23= 0.09on a 5MVA, 2.3kV base

(a) Determine the reactances X1X2X3 of the equivalent circuit on a 15MVA, 66kV base at the primary terminals. (b) Purely resistive loads of 7.5MW at 13.2kV and 5MW at 2.3kV are connected to the secondary and tertiary sides of the transformer, respectively. Draw the per-unit diagram, showing the per-unit impedances on a 15MVA, 66kV base at the primary terminals.

For an ideal transformer, the efficiency is

(a) 0%

(b) 100%

(c) 50%

A single-phase step-down transformer is rated 13MVA,66kV11.5kV. With the 11.5kVwinding short-circuited, rated current flows when the voltage applied to the primary is 5.5kV. The power input is read as 100 kW. DetermineReq1andXeq1 in ohms referred to the high-voltage winding.

Three single-phase transformers, each rated 10MW,66.4/12.5kV,60Hz, with an equivalent series reactance of 0.1 per unit divided equally between primary and secondary, are connected in a three-phase bank. The highvoltage windings are Y-connected and their terminals are directly connected to a 115kVthree-phase bus. The secondary terminals are all shorted together. Find the currents entering the high-voltage terminals and leaving the low-voltage terminals if the low-voltage windings are (a) Y-connected and (b) -connected.

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