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A 4 – pole, separately excited, wave wound DC machine with negligible armature resistance is rated for 230 V and 5 kW at a speed if 1200 rpm. If the same armature coils are reconnected to forms a lap winding, what is the rated voltage (in volts) and power (in kW) respectively at 1200 rpm of the reconnected machine if the field circuit is left unchanged?
1. 230 and 5
2. 115 and 5
3. 115 and 2.5
4. 230 and 2.5

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Correct Answer - Option 2 : 115 and 5

Wave wound

For wave wound, number of parallel pat (A1) = 2

∴ \(E_1=\frac{pϕ Nz}{60\ A_1}\)

⇒ \(E_1∝\frac{1}{A_1}\) {p, ϕ, N & z are constants}

Lap wound

for lap wound,

Number of parallel path = Number of poles

i.e., A2 = P = 4

∴ \(E_2∝\frac{1}{A_2}\)

So,

\(\frac{E_2}{E_1}=\frac{A_2}{A_1}\)

⇒ \(E_2=\frac{E_1}{2}\)

It is given that armature resistance (Ra) = 0

so, E ∝ V

⇒ \(\frac{E_2}{E_1}=\frac{V_1}{V_2}\)

⇒ \(V_2=\frac{E_2}{E_1}V_1\)

⇒ \(V_2=\frac{V_1}{2}\)

so, power (p) = V Ia

⇒ \(I_a=\frac{P}{V}\)

⇒ \(I_a\propto\frac{1}{V}\)

so,

\(\frac{I_{a_2}}{I_{a_1}}=\frac{V_1}{V_2}\ \{V_2=\frac{V_1}{2}\}\)

\(I_{a_2}=2I_{a_1}\)

The power of the machine will remain same as Plap = Pwave

The machine with lap wound armature coils will have rated voltage.

\(V_2=\frac{V_1}{2}=\frac{230}{2}=115\) V

∴ Plap = Pwave = p = 5 kW

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