A 4-pole, long-shunt, compound generator with a shunt field resistance of 100-ohm,
Series field resistance 0.04-ohm and an armature resistance of 0.02-ohm , has 378
wave–connected conductors in its armature. The flux per pole is 0.02 wb. If a load
resistance of 10-ohm is connected across the armature terminals and the generator is
driven at 1000 rpm, calculate the power absorbed by the load.

To calculate the power absorbed by the load in this compound generator, we'll follow these steps:

### Step 1: Calculate the Generated EMF (E)
The EMF generated by the generator can be determined using the formula:

\[
E = \frac{P \cdot Z \cdot \Phi \cdot N}{60 \cdot A}
\]

where:
- \( P \) = Number of poles = 4
- \( Z \) = Total number of armature conductors = 378
- \( \Phi \) = Flux per pole in weber = 0.02 Wb
- \( N \) = Speed in rpm = 1000 rpm
- \( A \) = Number of parallel paths in the armature = 2 (for wave-winding)

Substitute the values:

\[
E = \frac{4 \cdot 378 \cdot 0.02 \cdot 1000}{60 \cdot 2} = \frac{30240}{120} = 252 \text{ V}
\]

So, the generated EMF \( E \) is 252 V.

### Step 2: Calculate the Terminal Voltage (V)
Since the generator is long-shunt compound, the shunt field is connected across the armature terminals. The terminal voltage \( V \) is equal to the generated EMF minus the voltage drop in the armature and series field windings.

The total resistance in the armature and series field winding is:

\[
R_{\text{total}} = R_{\text{armature}} + R_{\text{series}} = 0.02 \, \Omega + 0.04 \, \Omega = 0.06 \, \Omega
\]

The current in the armature, \( I_a \), is the same as the current in the load because the load is connected across the armature terminals.

\[
I_a = \frac{V}{R_{\text{load}} + R_{\text{total}}} = \frac{252}{10 + 0.06} = \frac{252}{10.06} \approx 25.05 \, \text{A}
\]

Now, the terminal voltage \( V \) can be found using:

\[
V = E - I_a \cdot R_{\text{total}} = 252 - (25.05 \cdot 0.06) \approx 252 - 1.503 = 250.5 \, \text{V}
\]

### Step 3: Calculate the Power Absorbed by the Load
The power absorbed by the load is given by:

\[
P_{\text{load}} = \frac{V^2}{R_{\text{load}}}
\]

Substitute the values:

\[
P_{\text{load}} = \frac{250.5^2}{10} = \frac{62750.25}{10} = 6275.03 \, \text{W}
\]

### Final Answer:
The power absorbed by the load is approximately **6275 W** (or 6.28 kW).

---

Would you like more details on any step or have any questions?

Here are 5 related questions to consider:
1. How does the speed of the generator affect the generated EMF?
2. What is the effect of different load resistances on the power absorbed?
3. How do the field resistances impact the terminal voltage and current?
4. What is the difference between long-shunt and short-shunt configurations in compound generators?
5. How is the efficiency of this generator calculated?

**Tip:** Always check for voltage drops in the circuit when calculating terminal voltage, as they directly affect the output power.

A 4-pole, long-shunt, compound generator with a shunt field resistance of 100-ohm, Series field resistance 0.04-ohm and an armature resistance of 0.02-ohm, has 378 wave–connected conductors in its armature. The flux per pole is 0.02 wb. If a load resistance of 10-ohm is connected across the armature terminals and the generator is driven at 1000 rpm, calculate the power absorbed by the load.

Learn how to calculate the power absorbed by a load connected to a 4-pole, long-shunt compound generator. The problem involves calculating generated EMF, terminal voltage, and load power using key electrical engineering formulas. Suitable for undergraduate students in electrical engineering.

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