The derivation of maximum power transfer theorem for purely resistive DC circuits.
V_L= E x R_L/(R_L+r)
P_L= 〖V_L〗^2/R_L = (E^2 〖R_L〗^2/〖(R_L+r)〗^2 )/R_L = (E^2 R_L)/〖(R_L+r)〗^2 Calculating the most favourable value of the load resistor for maximum power transfer from a DC source with a internal resistance.
When any arbitrary load resistance may be selected.
Let R_L= kr ( k is a positive number)
P_L= kr/〖(kr+r)〗^2 E^2= k/〖(k+1)〗^2 E^2/r f = k/〖(k+1)〗^2 = 1/(k+2+k^(-1) ) k+k^(-1) ≥ 2 to find the least possible value of k+k^(-1)
To prove above: 〖(k-1)〗^2 ≥ 0 k^2-2k+1 ≥ 0 〖k+k〗^(-1) ≥ 2
Therefore k = 1 to minimize 〖k+k〗^(-1) which then maximizes P_L= …show more content…
The most favourable amount of internal resistance of a DC source for maximum power transfer to a load resistance.
Free to decide the amount of internal resistance of DC …show more content…
Which is reached by making r = 0.
Therefore P_(L_MAX )= E^2/R_L with an ideal source.
Thus for a completely resistive circuit the internal resistance of a DC source should be equal to zero for maximum power transfer to the load resistance.
The derivation of the maximum power transfer theorem for resistive and non-linear impedance AC circuits.
I – Phasor current
(ωL or-1/ωC)_L and (ωL or-1/ωC)_r= X_L and X_r respectively
I = (E_rms∠0°)/(Z+Z_r ) = (E_rms∠0°)/(r+R_L+j(X_r+X_L))
P_L= |I|^2 R_L
=R_L/(〖(r+R_L)〗^2+〖(X+X_L)〗^2 ) 〖E_rms〗^2
2.1) The maximum power transfer when we have the choice to decide both the load resistance and reactance.
To maximize P_L= R_L/(〖(r+R_L)〗^2+〖(X+X_L)〗^2 ) 〖E_rms〗^2, 〖(X+X_L)〗^2 must be minimized. This is carried out by selecting the load reactance to be the contrary of the AC source’s internal reactance i.e. X_L=X
Therefore load impedance should be the conjugate of the Thevenin impedance/ internal impedance of AC source for maximum power transfer to the load.
2.2) The maximum power formula when load reactance can’t be altered, but the load resistance can be
V_L= E x R_L/(R_L+r)
P_L= 〖V_L〗^2/R_L = (E^2 〖R_L〗^2/〖(R_L+r)〗^2 )/R_L = (E^2 R_L)/〖(R_L+r)〗^2 Calculating the most favourable value of the load resistor for maximum power transfer from a DC source with a internal resistance.
When any arbitrary load resistance may be selected.
Let R_L= kr ( k is a positive number)
P_L= kr/〖(kr+r)〗^2 E^2= k/〖(k+1)〗^2 E^2/r f = k/〖(k+1)〗^2 = 1/(k+2+k^(-1) ) k+k^(-1) ≥ 2 to find the least possible value of k+k^(-1)
To prove above: 〖(k-1)〗^2 ≥ 0 k^2-2k+1 ≥ 0 〖k+k〗^(-1) ≥ 2
Therefore k = 1 to minimize 〖k+k〗^(-1) which then maximizes P_L= …show more content…
The most favourable amount of internal resistance of a DC source for maximum power transfer to a load resistance.
Free to decide the amount of internal resistance of DC …show more content…
Which is reached by making r = 0.
Therefore P_(L_MAX )= E^2/R_L with an ideal source.
Thus for a completely resistive circuit the internal resistance of a DC source should be equal to zero for maximum power transfer to the load resistance.
The derivation of the maximum power transfer theorem for resistive and non-linear impedance AC circuits.
I – Phasor current
(ωL or-1/ωC)_L and (ωL or-1/ωC)_r= X_L and X_r respectively
I = (E_rms∠0°)/(Z+Z_r ) = (E_rms∠0°)/(r+R_L+j(X_r+X_L))
P_L= |I|^2 R_L
=R_L/(〖(r+R_L)〗^2+〖(X+X_L)〗^2 ) 〖E_rms〗^2
2.1) The maximum power transfer when we have the choice to decide both the load resistance and reactance.
To maximize P_L= R_L/(〖(r+R_L)〗^2+〖(X+X_L)〗^2 ) 〖E_rms〗^2, 〖(X+X_L)〗^2 must be minimized. This is carried out by selecting the load reactance to be the contrary of the AC source’s internal reactance i.e. X_L=X
Therefore load impedance should be the conjugate of the Thevenin impedance/ internal impedance of AC source for maximum power transfer to the load.
2.2) The maximum power formula when load reactance can’t be altered, but the load resistance can be