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9.1. INSTANTANEOUS AND AVERAGE POWER
 The instantaneous power is the power at any instant of
time. It is the rate at which an element absorbs energy.
 Let the voltage and current at the terminals of the circuit
be;
9.1. INSTANTANEOUS AND AVERAGE POWER
 𝑉𝑚 𝑎𝑛𝑑 𝐼𝑚 are the amplitudes (or peak) values, 𝜃𝑣 and 𝜃𝑖
are the phase angles of the voltage and current.
 The instantantaneous power absorbed by the circuit is;
9.1. INSTANTANEOUS AND AVERAGE POWER
 This shows us that the instantaneous power has two parts.
 The first part is constant or time independent. İts value
depends on the phase on difference between the voltage
and the current.
 The second part is a sinusoidal function whose frequency
is 2ω, which is twice the angular frequency of the voltage
or current.
9.1. INSTANTANEOUS AND AVERAGE POWER
 In figure; 𝑇 = 2𝜋/𝜔 is the period of voltage or current.
 p(t) is periodic. 𝑝 𝑡 = 𝑝(𝑡 + 𝑇0 ) and has a period of 𝑇0 =
𝑇
2
 We also observe that p(t) is positive for some part of each cycle and negative
for the rest of cycle
 When p(t) is positive, power is absorbed by the circuit.
 When p(t) is negative power is absorbed by the source; thar is power is
transferred from the circuit to the source. This is possible because of the
storage elements in the circuit.
9.1. INSTANTANEOUS AND AVERAGE POWER
9.1. INSTANTANEOUS AND AVERAGE POWER
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9.1. INSTANTANEOUS AND AVERAGE POWER
9.1. INSTANTANEOUS AND AVERAGE POWER
Example 9.1:
For given voltage and current, find the intantaneous power
and the average power absorbed by the passive linear
network.
Solution:
Example 9.1:
Example 9.2:
Solution:
Example 9.3:
For the circuit in Figure find the average
power supplied by the source and the
average power absorbed by the resistor.
Solution:
Example 9.3:
Example 9.4:
Determine the power generated by
each source and the average power
absorbed by each passive element in
the circuit of figure.
Solution:
Example 9.4:
This average power absorbed by the source, in view of the direction 𝐼2 and
the polarity of the voltage source. That is the circuit delivering average
power to the voltage source.
Example 9.4:
Example 9.4:
9.2. Maximum Average Power Transfer
9.2. Maximum Average Power Transfer
Our objective is to adjust the load parameters 𝑅𝐿 and 𝑋𝐿 so
that P is maximum.
For maximum average power transfer, 𝑍𝐿 must be selected
so that 𝑋𝐿 = −𝑋𝑇ℎ and 𝑅𝐿 = 𝑅𝑇ℎ
9.2. Maximum Average Power Transfer
In a situation in whcih the load is purely real, the condition for the
maximum power transfer is obtained by setting, 𝑋𝐿 = 0.
Example 9.5:
Determine the load impedance 𝑍𝐿 that
maximizes the average power drawn
from the circuit of Figure. What is the
maximum average power?
Solution:
Example 9.5:
Example 9.6:
Solution:
Example 9.6: