00:02
Hello, in the question we have given this circuit.
00:04
So, in the circuit diagram we have given the value of this resistors r1, r2 and r3 and the potential, the battery is given the emf of battery.
00:15
So, the voltage of battery is 10 v.
00:18
Now, we have to find out what is the potential drop across this r2, the resistance r2.
00:23
So, now we can find that this, these two are in parallel.
00:29
So, if we find out the equivalent, if i want to find out the equivalent, what i will do first is i will compute the parallel resistance and add r1 to it because r1 is in series to this both.
00:43
So, 1 upon rp will be equal to 1 upon 5 .6 plus 1 upon 4.
00:50
So, this gives me 4 times 5 .6, sorry, so this gives me 4 plus 5 .6 divided by 5 .6 times 4.
01:03
So, rp will be equal to 5 .6 times 4 divided by 4 plus 5 .6.
01:10
So, this will be 5 point, so this turns out to be 2 .33 ohms.
01:20
Now, this rp that is the parallel resistance and this rs, r1 is in, so r equivalent will be equal to r1 because these two are in series.
01:31
So, this will be 1 plus 2 .33.
01:35
So, this r equivalent then turns out to be 3 .33 ohms.
01:42
So, this is the r equivalent.
01:44
Now, from this r equivalent, we will first find out how much current is flowing.
01:49
So, this circuit will look somewhat like this then.
01:55
So, we will have one resistor over here...