Question
Download Solution PDFIn the circuit, the maximum power that can be transferred to Load ZL is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
The given circuit is
\(\rm P_{ZL}=I^2_{rms}.Z_L\)
\(\rm I_{rms}=\frac{I_m}{√2}\)
Zs = Rs + j ωLs = 10 + j 1000 × 10 × 10-3 = 10 + j 10 Ω
\(\rm I_m=\frac{i(t)(10+j10)}{10+j10+Z_L}\)
As we know ZL = Zs* = 10 - j10 = RL + j XL
|ZL| = 10√2 Ω
\(\rm I_m=\frac{10√2(10+j10)}{10+j10+10-j10}\)
\(|I_m|=\frac{10√2×10√2}{20}=10\ \rm Amp\)
\(\rm I_{rms}=\frac{I_m}{√2}=\frac{10}{√2}\)
\(\rm P_{Z_L}=I^2_{rms}.Z_L=\left(\frac{10}{√2}\right)^2×10\)
PZL = 500 W
Important Points Maximum power transfer theorem for AC circuits:
The maximum power transfer theorem states that the maximum power flow through an AC circuit will occur when the load impedance is equal to the complex conjugate of the source impedance.
ZL = ZS*
|ZL| = |ZS|
ZL = Load impedance
ZS = Source impedance
Important points:
|
Load variable |
The load impedance for maximum power transfer |
|---|---|
|
RL and XL are variable |
RL = RS XL = -XS ZL = ZS* |
|
RL only varied and XL = Constant |
RL =√(RS2 + (XL + XS)2) |
|
RL only varied and XL = 0 |
RL =√(RS2 + XS2) |
Last updated on Mar 26, 2025
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