00:01
For the superposition theorem we will consider only 90v volt.
00:07
So we will make the 6 ampere as the open circuit and then 40 volts will be as short circuit.
00:29
By doing this we will get this will be the 90 volt this will be the 90 volt this will be the 30 this is 60 this is 10 this is also 30 and this will be 20 we can write this will be 30 60 10 30 20 if we reduce the circuit then if we take the equivalent of this this will be 30 in the equivalence of this, this will become 16 .097.
01:15
Again, this is minus plus.
01:19
So this will be taken as 30 oom and the 90.
01:27
So we will write this.
01:30
The voltage along the 16 .097 resistance can be given by 90, multiply by 16.
01:39
16 .097 as 13 and 16 .097 are in series divided by 16 .097 plus 30.
01:52
So we will get 31 .427 volt.
01:58
Now further we can increase the circuit.
02:03
So we will find the voltage across the 10 oom for this we can write this 31 .427 volt multiply by 10 divided by 10 plus 12.
02:22
So this will be equal to v1 we will get 14 .285 volt.
02:31
Now the next thing we will consider only 40 volt for this.
02:40
We will take 90 volt as the short circuit and the 60 ampere source as an open circuit...