1.

In the question above, the value of equivalent voltage source is?

A. V<sup>‘</sup>=(V<sub>1</sub>G<sub>1</sub>+V<sub>2</sub>G<sub>2</sub>+⋯.+V<sub>n</sub>G<sub>n</sub>)
B. V<sup>‘</sup>=((V<sub>1</sub>G<sub>1</sub>+V<sub>2</sub>G<sub>2</sub>+⋯.+V<sub>n</sub>G<sub>n</sub>))/((1/G<sub>1</sub>+1/G<sub>2</sub>+⋯1/G<sub>n</sub>))
C. V<sup>‘</sup>=((V<sub>1</sub>G<sub>1</sub>+V<sub>2</sub>G<sub>2</sub>+⋯.+V<sub>n</sub>G<sub>n</sub>))/(G<sub>1</sub>+G<sub>2</sub>+⋯G<sub>n</sub>)
D. V<sup>‘</sup>=((V<sub>1</sub>/G<sub>1</sub>+V<sub>2</sub>/G<sub>2</sub>+⋯.+V<sub>n</sub>/G<sub>n</sub>))/( G<sub>1</sub>+G<sub>2</sub>+⋯G<sub>n</sub>)
Answer» D. V<sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>=((V<sub>1</sub>/G<sub>1</sub>+V<sub>2</sub>/G<sub>2</sub>+‚Äö√Ñ√∂‚àö¬£‚àö√≤.+V<sub>n</sub>/G<sub>n</sub>))/( G<sub>1</sub>+G<sub>2</sub>+‚Äö√Ñ√∂‚àö¬£‚àö√≤G<sub>n</sub>)


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