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The internal impedance of a source is 3 + j4 Ω, it is desired to supply maximum power to a resistive load. The load resistance should be
- A3 Ω
- B4 Ω
- C7 Ω
- DNone of these
Answer not set
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CCommunity
According to maximum power transfer theorem in the AC circuit, to draw maximum power the load impedance must be complex conjugate of source circuit internal impedance.
Here purely resistive load is chosen, hence for maximum average power transfer, the load resistance is equal to the magnitude of source internal impedance.
$({R_L} = \;\sqrt {{R^2} + {X^2}} = \;\sqrt {{3^2} + {4^2}} = \;5\;{\rm{\Omega }})$
Here purely resistive load is chosen, hence for maximum average power transfer, the load resistance is equal to the magnitude of source internal impedance.
$({R_L} = \;\sqrt {{R^2} + {X^2}} = \;\sqrt {{3^2} + {4^2}} = \;5\;{\rm{\Omega }})$