Instanttaneous Power And Average Power Flowing To A One-Port Network
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Instanttaneous Power And Average Power Flowing To A One-Port Network

Instantaneous Power Average Power

(a) What is a one-port network?
(b) What is a two-port network?
(c) A one-port network with passive sign convention has v = Vmcos ωt V and i = I1 + I2cos (ωt + θ) A
Determine: (i) the instantaneous power flowing to the the network; (ii) the average power to the network.

Instantaneous Power Average Power

The strings: S7P3A32 (force - push).

The math:
Pj Problem of Interest is of type force (push).

(a) A pair of terminals constitute a port: if there is an identifiable voltage across the terminals and the current into one terminal is the same as the current out of the other terminal. is
Figure 1.1(a) consists of two one-port networks (A and B). Network, A is a one-port network when viewed from from the left of terminals 1, 2. Network B is a one-port network when viewed from the right side of terminals 1,2.

(b) Figure 1.1(b) is a two-port network if currents I1in and I1out are equal; and currents I2in and I2out are equal. The four variables V1, V2, I1 and I2 (only two of which can be independent) characterize the network.

(ci) A one-port network with a passive sign convention implies that current is directed from the positive to the negative terminal.
Instantaneous power = p(t) = v(t)i(t) = Vmcos ωt[I1 + I2cos (ωt + θ)] = VmI1cos ωt + VmI2cos ωtcos (ωt + θ).
VmI2cos ωtcos (ωt + θ) = VmI2[cos 2(ωt + θ) + cosθ]; base on the product-sum trigonometric identity;
Cos A cos B = 1/2[cos (A - B) + cos (A - B)].

(cii) Average power Po = 1/(2π)[(∫ p(t)dt) t=2π - (∫ p(t)dt) t=0]
After expansion of the trigonometric function and integration:
Po = (1/2)VmImcos θ.


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