A power transmission system is modeled as shown in Fig. 9.82...

A power transmission system is modeled as shown in Fig. $9.82 .$ Given the source voltage $\mathbf{V}_{s}=115 / 0^{\circ} \mathrm{V},$ source impedance $\mathbf{Z}_{s}=1+j 0.5 \Omega,$ line impedance $\mathbf{Z}_{\ell}=0.4+j 0.3 \Omega,$ and load impedance $\mathbf{Z}_{L}=23.2+j 18.9 \Omega,$ find the load current $\mathbf{I}_{L}$.

A power transmission system is modeled as shown in Fig. $9.82 .$ Given the source voltage $\mathbf{V}_{s}=115 / 0^{\circ} \mathrm{V},$ source impedance
$\mathbf{Z}_{s}=1+j 0.5 \Omega,$ line impedance
$\mathbf{Z}_{\ell}=0.4+j 0.3 \Omega,$ and load impedance $\mathbf{Z}_{L}=23.2+j 18.9 \Omega,$ find the load current $\mathbf{I}_{L}$.

Key Concepts

Phasor Representation

Phasor representation is a method used in AC circuit analysis where sinusoidal signals are represented as complex numbers. This approach transforms differential equations into algebraic equations, facilitating operations such as addition and multiplication that account for both magnitude and phase of the signals.

Complex Impedance

Complex impedance extends the concept of resistance to AC circuits by combining both resistive and reactive (inductive and capacitive) elements into one complex number. This allows for the analysis of circuits where current and voltage are out of phase, making it possible to apply techniques like series impedance summation.

Series Circuit Analysis

Series circuit analysis involves summing impedances in a sequential manner along a transmission line or network. This technique simplifies the computation of total impedance seen by the source, which is critical when determining current flow and voltage drops throughout an AC circuit.

AC Ohm's Law

AC Ohm's Law generalizes the conventional Ohm's law to alternating current circuits by incorporating complex impedance. It states that the phasor current is obtained by dividing the phasor voltage by the total complex impedance, thus accounting for both magnitude and phase shifts in the circuit.

Transcript

00:01 Hello student, in this problem a model of transmission line is given and where the value of source voltage vs equals to 1115 angle of 0 degree volt and the value of series impedance z s equals to 1 plus iota 0 .5 oom and the value of line impedance zl equals to z zl equals to 0 .4 plus iota 0 .3 and the value of load impedance, zl equals to 23 .2 plus iota 18 .9 oar.

00:55 Okay.

00:56 Now we have to find out the value of, value of load current il.

01:01 We have to find out value of the load current il.

01:04 So, we know that the load current is given as il equals to, il equals to vs upon z, where z is the total impedance, total impedance of line, total impedance of line.

01:26 So first we have to calculate the total impedance of the line, okay? so, z equals to, z equals to higher series impedance is zs, resource impedance is zs, zs, plus line impedance zl plus load impedance zl plus again series line impedance zl then effective impedance z equals to z s plus 2 zl plus load impedance zl okay so now put the value in this equation z equals to z s equals to 1 plus iota 0 .5 om plus zl is 2, 0 .0 .0 .5 plus zl equals to 2, line impedance is 0 .4, plus iota 0 .3 om, plus load impedance is 23 .2 plus iota 18 .9 arm.

02:30 Okay? so the effective value of load impedance, total impedence, sorry, total impedance z equals to, total impedance z equals to 1 plus iota 0 .5 plus 0 .8 plus iota 0 .6 plus 23 .2 plus iota 18 .9.

02:57 Okay, so now total impedance z equals to sum of the real value 1 plus 0 .8.

03:04 Plus 23 .2 so the effective value is 25 plus and the submission of this part okay submission of this part so it is it is iota 20 so this is the complex form so if we convert it in polar form then the value will be 32 .0 2 angle of 38 .66 degree okay the image impedance in polar form can be calculated by using the formula a plus iota b equals to square root of a square plus b square...

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