The armature of a four-pole D C shunt generator has 378 wave connected conductors. The armature and shunt winding resistance of the generator is 1 ohm and 100 ohm respectively. 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.
Question 4.42: The armature of a four-pole D C shunt generator has 378 wave...
- Number of wave connected conductors (N) = 378
- Armature winding resistance (Ra) = 1 ohm
- Shunt winding resistance (Rsh) = 100 ohm
- Flux per pole (Φ) = 0.02 Wb
- Load resistance (Rload) = 10 ohm
- Generator speed (N) = 1000 rpm
Step 1:
Find the generated emf (Eg) using the formula Eg = (P Z φ N) / (60 A), where P is the number of poles, Z is the total number of conductors, φ is the flux per pole in Weber, N is the speed of the machine in rpm, and A is the number of parallel paths.
Step 2:
Calculate the line current (IL) using the formula IL = V / RL, where V is the terminal voltage and RL is the load resistance.
Step 3:
Determine the shunt field current (Ish) using the formula Ish = V / Rsh, where Rsh is the shunt field resistance.
Step 4:
Calculate the armature current (Ia) by adding the line current and shunt field current, i.e., Ia = IL + Ish.
Step 5:
Calculate the armature current (Ia) by adding the line current and shunt field current, i.e., Ia = IL + Ish.
Step 6:
Substitute the value of V into the formula IL = V / RL to find the load current (IL).
Step 7:
Finally, calculate the power absorbed by the load (P) using the formula P = V * IL.
By following these steps, we can find the values of all the required parameters.
Final Answer
The conventional circuit is shown in Fig. 4.65.
\text { Generated emf, } E_{g}=\frac{P Z \phi N}{60 A}=\frac{4 \times 378 \times 0 \cdot 02 \times 1000}{60 \times 2}=252 VLine current, I_{L}=\frac{V}{R_{L}}=\frac{V}{10} (where V is terminal voltage)
Shunt field current, \quad I_{s h}=\frac{V}{R_{s h}}=\frac{V}{100}
Armature current,
I_{a}=I_{L}+I_{s h}=\frac{V}{10}+\frac{V}{100}=(0 \cdot 11 V)Using the relation, \quad E_{g}=V+I_{a} R_{a} ;
252=V+0.11 V \times 1 \cdot 0\therefore \quad Terminal voltage, \quad V=227 volt
\therefore \text { Load current, } I_{L}=\frac{V}{R_{L}}=\frac{227}{10}=22 \cdot 7 APower absorbed by the load, P=V I_{L}=227 \times 22.7= 5 \cdot 1 5 3 k W
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