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SSC JE Electrical 2019 with solution SET-2
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Observe the given figure. Find Thevenin’s resistance as seen from open-circuited terminals?

To find the Thevenin’s resistance the current source is open-circuited and the voltage source is short-circuited.

SSC JE Electrical 2019 with solution SET-2
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In a 4 pole. 20 kW, 200 V wave wound DC shunt generator, the current in each parallel path will be

Given

Power (P) = 20 kW = 20 × 103 W

Voltage (V) = 200 V

P = VI = I = P/V

I = (20 × 103)/200

I = 100A

For wave wound No. of Parallel Path = 2

Current in Each parallel Path for wave wound

I = 100/2 = 50 A

SSC JE Electrical 2019 with solution SET-2
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In an alternator, if α = 60°, then find the pitch factor?

The factor by which, the induced E.M.F gets reduced due to short pitching is called pitch factor or coil span factor denoted by Kc. It is given as

Kc = cosα/2

given

α = 60°

Kc = cos60/2

Kc = cos30°

Kc = √3/2

SSC JE Electrical 2019 with solution SET-2
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Observe the given network and answer the question? The potential at point a = ?

For the given network the potential at point a can be found by applying KVL.

For loop 1

Va = E1 − I1R1

Va = I3R3

For loop 2

Va = E2 − I2R2

SSC JE Electrical 2019 with solution SET-2
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A series magnetic circuit will have

In the same way that we can have series, parallel or series-parallel electric circuits, we can also have equivalent series and series-parallel (but not parallel) magnetic circuits.

In practice, a magnetic circuit may consist of several parts in a series of different lengths, cross-sectional areas, and permeabilities. In such a series magnetic circuits, all the reluctances of several parts will get summed up together (as resistors in an electric circuit) to form the net reluctance of the circuit.

In series magnetic circuits the flux passing through each part will be the same (as current in the series electric circuits).