# A differentially compound DC generator has its series and shunt field providing 20% and 80% flux respectively. The series field is suddenly short circ

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A differentially compound DC generator has its series and shunt field providing 20% and 80% flux respectively. The series field is suddenly short circuited without changing the load. Then the terminal voltage ignoring all the resistances
1. Remains same
2. Reduce by 20%
3. Increases by 33.33%
4. Reduces by 80%

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Correct Answer - Option 3 : Increases by 33.33%

In a differentially compound DC generator,

Flux due to shunt field ϕrh = 80% of ϕ

Flux due to series field ϕre = 20% of ϕ

Since all the resistances are ignored, the voltage drop is zero.

⇒ Vt = Eg

As, Vt = Eg - IaRa

Vt = terminal voltage

Eg = generated voltage

The generated voltage is,

${{E}_{g}}=\frac{Nϕ ZP}{60A}$

⇒ Eg ∝ ϕ

⇒ Vt ∝ ϕ ⇒ Vt ∝ (ϕrh – ϕre)

$⇒ {{V}_{{{t}_{old}}}}\propto 0.8ϕ -0.2ϕ ⇒ {{V}_{{{t}_{old}}}}\propto 0.6ϕ$

Now, the series field is suddenly short-circuited

⇒ ϕre = 0

$⇒ {{V}_{{{t}_{new}}}}\propto ϕ ⇒ {{V}_{{{t}_{new}}}}\propto \left( {{ϕ }_{rh}}-{{ϕ }_{re}} \right)$

$⇒ {{V}_{{{t}_{new}}}}\propto 0.8ϕ$

Percentage change in terminal voltage = 100 ×  (Vtnew - Vtold) / Vtold

⇒ % change in Vt = 100 × (0.8 ϕ - 0.6 ϕ) / 0.6 ϕ

⇒ % change in Vt = 100 × 1 / 3 = 33.33%

∴ Terminal voltage increases by 33.33%