Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.
<u>Given the following data:</u>
- Number of armature conductors = 144 slots
- Number of poles = 4 poles
- Number of parallel paths = 2
To find the magnetic flux per pole:
Mathematically, the emf generated by a DC generator is given by the formula;
×
<u>Where:</u>
- E is the electromotive force in the DC generator.
- Z is the total number of armature conductors.
- N is the speed or armature rotation in r.p.m.
- P is the number of poles.
- A is the number of parallel paths in armature.
First of all, we would determine the total number of armature conductors:
× ×
Z = 864
Substituting the given parameters into the formula, we have;
×
×
<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>
Therefore, the magnetic flux per pole is 0.0274 Weber.
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Answer:
False.
Explanation:
Welding due to the thermal stress caused by the piece and large temperature changes can cause residual stresses in the material due to the efforts that are manifested in it. These efforts can cause microfractures, weakened areas and also a stress and changes in the crystalline structure that does not allow it to maintain the initial properties of the material and be prone to subsequent failures and corrosion.
Answer:
From register to register next is register to indexed storage and next to register to storage and the to single oprand and finally to storage to storage.
Explanation:
Within the CPU, the most important components are registers. Data may be moved between registers, may be added or subtracted from the current contents of a register, and can be shifted or rotated within a register or between registers. Each instruction in the instruction set is executed by performing these simple operations, using the appropriate choice of registers and operations in the correct sequence for the particular instruction.
Answer:
domestic, public, commercial, and industrial uses.
Answer:
R = 1 kΩ
i_L = 2.2 mA
Explanation:
The complete question is given in attachment:
Given:
- The gain i_L / i_i = 11
- Given circuit Attachment
Find:
- Find the required value of R
- If the amplifier is fed with a current source having a current of 0.2 mA and a source resistance of 10 kΩ, find iL
Solution:
- Use KVL on right most loop (load loop):
v_x = 0 - i_i*10 = -10i_i
- Using ohm's law:
i = (0 - v_x) / R
i = 10*i_i / R
- Use KCL at the node:
i_L = i + i_i
- Substitute the the two i:
i_L = 10*i_i / R + i_i
i_L / i_i = ( 10/ R + 1 )
11 = ( 10/ R + 1 )
10 / R = 10
R = 1 kΩ
- The input resistance = 0 because there is a virtual ground at input. Then the value of source resistance will have no effect on resulting i_l, hence:
i_L = i_L / i_i * i_i
i_L = 11*0.2
i_L = 2.2 mA