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