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

**(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 = V_{m}cos ωt V and i = I_{1} + I_{2}cos (ωt + θ) A

Determine: (i) the instantaneous power flowing to the the network; (ii) the average power to the network.

**The strings**:
S_{7}P_{3}A_{32} (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 I_{1in} and I_{1out} are equal; and currents I_{2in} and I_{2out} are equal. The four variables V_{1}, V_{2}, I_{1} and I_{2} (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) = V_{m}cos ωt[I_{1} + I_{2}cos (ωt + θ)] = V_{m}I_{1}cos ωt + V_{m}I_{2}cos ωtcos (ωt + θ).

V_{m}I_{2}cos ωtcos (ωt + θ) = V_{m}I_{2}[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 P_{o} = 1/(2π)[(∫ p(t)dt) ^{t=2π} - (∫ p(t)dt) _{t=0}]

After expansion of the trigonometric function and integration:

P_{o} = (1/2)V_{m}I_{m}cos θ.

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The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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