00:01
Alright, so let's find the thavenines and norton's equivalent circuits for the following circuit.
00:07
So let's find the open circuit voltage voc is an open circuit voltage and the short circuit current.
00:14
Voc is the voltage across the terminals a and b when under open circuit conditions.
00:22
Alright, so under open circuit conditions what happens is the current will flow only through this path.
00:31
So that means these two resistors will be in series right so i can i can convert this circuit something like this so this is a 15 -volt source and this is the 5 -ooms source and this is the dependent source so from here 0 .5 ix of the current is going and this is ix and then the 10 oms and 10 oms i can combine to 20 oms and this is our terminals a and b right so obviously when you apply kitchchof's voltage law so let's see how much current will i mean how much voltage across a b gets dropped now since i x is coming from 5 oms and point 5 i x is went to this so obviously 0 .5 ix will be distributed along this resistor and let's apply kchof's voltage law to the bigger look since we can't apply kvl for the dependent loop containing the dependent so we should apply the kitchhaw's voltage law along this.
01:33
So if you apply it will be 15 minus 5 into ix minus 0 .5 x into 20 is equal to 0.
01:44
So 15 is equal to 5 x plus 0 .5 into 20 that is 10 x so when you have 5 x and 10 x you'll be getting 15 is equal to 15 x so basically 15 x is equal to 15 so ix is equal to 1 amp 1 amp you right so one ampere means so that means what will the current flowing into this resistor point five amps so that means it will be 0 .5 amps so 0 .5 ams means what will the voltage drop 20 into 0 .5 is 10 volts so open circuit voltage so voc is equal to 10 volts now let's analyze the circuit in the short circuit conditions so in short circuit conditions we should open out the dependent sources so we have 15 volts and we have here how much we have here five oms with an i x so five oms with the current i so no more i x is required so i will simply write i and this is open circuit and now ten oms ten oms so we have ten oms we have another ten oms but now the circuit is short -circuited at the terminals a and b so that means definitely the current i will not pass through this resistor it will directly choose the path of this short short circuit, all right? because currents always like short secutes.
03:07
So that means when you apply the kitchoffs voltage law, it will be 15 minus 5 plus 10, that is 15 i is equal to 0...