CALCU-solve it5. Let |V_1|=1.05,|~V 2|=1, z_ line

solve it

Solution by Steps

step 1

We start with the expression for the complex power

s_{12}

given by

s_{12} = \frac{|V_1|^2}{Z_{line}}

where

Z_{line} = 0.6 \angle 80^\circ

. We need to find

\theta_{12}

such that

s_{12}

is purely reactive

step 2

The condition for

s_{12}

to be purely reactive is that the real part must be zero. This can be expressed as

\text{Re}(s_{12}) = 0

. We can express

|V_1|

and

|V_2|

in terms of their magnitudes and angles:

|V_1| = 1.05

and

|V_2| = 1

step 3

The impedance

Z_{line}

can be expressed in rectangular form as

Z_{line} = 0.6 (\cos(80^\circ) + j \sin(80^\circ))

. We calculate

\cos(80^\circ)

and

\sin(80^\circ)

to find the real and imaginary parts

step 4

Setting the real part of

s_{12}

to zero gives us the equation

\frac{1.05^2 \cos(\theta_{12})}{0.6 \cos(80^\circ)} = 0

. This implies

\cos(\theta_{12}) = 0

, leading to

\theta_{12} = \frac{\pi}{2}

or

\theta_{12} = \frac{3\pi}{2}

. Since 0 < \theta_{12} < 2\pi , we take

\theta_{12} = \frac{\pi}{2}

Answer

\theta_{12} = \frac{\pi}{2}

Key Concept

The concept of complex power in AC circuits and the conditions for it to be purely reactive.

Explanation

For

s_{12}

to be purely reactive, the real part must equal zero, which leads us to find specific angles that satisfy this condition. In this case,

\theta_{12} = \frac{\pi}{2}

meets the requirement.

Solution by Steps

step 1

The buck converter operates in two modes in Discontinuous Conduction Mode (DCM): the "On" mode when the switch is closed and the "Off" mode when the switch is open

step 2

The equivalent circuit for the "On" mode consists of the voltage source

V_g

, the inductor

L

, and the switch

SW

closed, while the diode

D1

is reverse-biased. In the "Off" mode, the switch

SW

is open, and the inductor

L

discharges through the diode

D1

to the capacitor

C

and the load resistor

R

step 3

To derive the input to output DC gain

\left(\frac{V_g}{V_o}\right)

, we can use the relationship:

V_o = D \cdot V_g

, where

D

is the duty cycle. Thus, the gain is given by

\frac{V_g}{V_o} = \frac{1}{D}

step 4

The criteria for the converter to operate in DCM is that the inductor current must reach zero during the switching cycle. This occurs when the inductor current

i_L

is less than the average load current

I_{avg}

, which can be expressed as I_{avg} < \frac{V_o}{R}

step 5

The time waveforms of the duty cycle

D

, switch voltage

SW

, inductor current

i_L

, and capacitor current

i_C

can be sketched. The peak value of

i_L

occurs at the end of the "On" mode, while the valley value occurs at the end of the "Off" mode. The capacitor current

i_C

will be a function of the load and the inductor current

Answer

The buck converter operates in two modes in DCM: "On" and "Off". The input to output DC gain is

\frac{1}{D}

, and the criteria for DCM operation is I_{avg} < \frac{V_o}{R}.

Key Concept

Buck converters can operate in different modes depending on the current through the inductor and the duty cycle.

Explanation

Understanding the modes of operation and the relationships between the circuit components is crucial for analyzing buck converters in DCM.

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