Why dont you reply to the doubts listed in the comment section? This is the only drawback that you have. SO please reply to the doubts listed in the comments!!!
Remember in Thevenin's theorem how we used to calculate the Rth? When we're finding the equivalent resistance of the circuit, we should deactivate the sources....i.e., short circuit the voltage source and open circuit the current source. Same thing here. We should deactivate the sources to find input impedance and output impedance.
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Bro , u r not earning views u r earning blessings ❤️❤️
Never ever seen as lecture videos like yuhh❤
What a explanation like wow❤
Thank you sir ❤️
Why dont you reply to the doubts listed in the comment section? This is the only drawback that you have. SO please reply to the doubts listed in the comments!!!
6 years completed now also no response
How did you equate, Vs as zero,in the case of at apply KVL to the input loop, plz explain sir,
bcz Vs is to be short circuited meaning Vs = 0
hfe= 30ohms, hie=1000ohms,RL= 1000ohms , how to find voltage gain and current gain.
Thank you so much, I will pray for you and your family
Well explained sir tq so much
why do we need to short circuit input source &open circuit for output terminal ?? why cant it be done as earlier cases
Arrey Wah! xD
why hrv2 is negative while using kvl in clock wise manner?
refer to the hybrid equation: V1 = h1i1 + hrv2
If RL= infinity, i2=0, Zo=v2/i2=infinity?
Thanks sir❤
Sir Zo=AiRL/(hf+hoAiRL)
Sir op!!
At 4:45 the signs used in KVL is confusing😕❓ I am getting -i1Rs+i1hi+hrv2=0... Please advice
if we short ckt the input Vs then why current i1 is not zero??? plz explain sir
current will be zero for open ckt..whereas voltage will be zero for short ckt
then...from where does this i1 come from ....
Why is Vs kept zero? while taking KVL?
did you find its answer??
Remember in Thevenin's theorem how we used to calculate the Rth? When we're finding the equivalent resistance of the circuit, we should deactivate the sources....i.e., short circuit the voltage source and open circuit the current source. Same thing here. We should deactivate the sources to find input impedance and output impedance.
Why always ( ho*hi-hf*hr )taken as Δh? why not any other subtraction containing h parameters taken as Δh
At last my brain be like :- കട മിഴിയിൽ, കമലദലം ,ഹേയ് 😂🤣
Not funny
While finding output resistance..why shoukd we consider vs =0
Why (i1 Rs) is negative in KVL while calculating output impedance?
he is calculating in direction of current (that is clockwise) ,so voltage drop in resister(Rs) is --i1Rs
why current gain is not written as hf
After dived on both side i1 how will - convert to +
In This video we can seen in 1:56 resume see it and pls reply me
Hi. Great vid, but your 1 sometimes looks like a 2. It's kinda annoying when you're taking notes
sir, actually Zo=V2/I2,but V2=-I2RL then Zo=-RL
will u please explain it?
As il=-i2 ..here at output loop il=0 because we open circuited the RL no current flow so iL=0 .so we cannot take i2 becomes zero. I2=0 then Zo=0..
🙄 first define these things ....1.input impedance,2.input resistance,3.input impedance parameter .......And output impedances.....
is output impedance same for all configuration
please provide pdf for h parameters topic
Where we get the Ai ?
Current gain
Saptarshi re juta dia pitamu , dhishum dhishum
How can will come occur rL into 1/rl pls explain
Thank you sir...but how we get Ai?
hi guys
output impedance shud be V2/iL i.e. -V2/i2??
Oi saptarshi shuor
H parameters of Both are same current gain and voltage gain in CC , CB,CE please tells me 😢