Fig. shows two alternators with identical speed / load characteristics connected in parallel. The circulating current under no-load condition is

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  1. \(\frac{(E_1-E_2)}{(Z_1+Z_2)}\)
  2. \(\frac{(E_1+E_2)}{(Z_1+Z_2)}\)
  3. \(\frac{(Z_1Z_2)}{(Z_1+Z_2)}(E_1-E_2)\)
  4. \(\frac{(Z_1Z_2)}{(Z_1-Z_2)}(E_1-E_2)\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{(E_1-E_2)}{(Z_1+Z_2)}\)
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  • Two identical alternators have impedances Z1 and Zconnected in parallel produces, E& Eas their individual induced emf’s.
  • If no load is connected between the alternators, the circulating current will flow in the circuit as given by \(\frac{{{E_1} - {E_2}}}{{{Z_1}\; + \;{Z_2}}}\)
  • Here, impedances are different also it should be noted that E& Eare in phase relative to the external circuit but are in direct phase opposition in the local circuit, when it is in parallel.

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