00:01
Hi, in the given problem emf of the battery is given as e is equal to 24 .0 volt.
00:15
Its internal resistance is given as r is equal to 3 .00 oom.
00:28
But external resistance in the circuit means r that is missing and we have to find its value provided.
00:41
The power dissipated across r is given as p is equal to 21 .0 watt.
00:59
So using these parameters, we have to find two possible values of the resistance are.
01:07
So first of all, the net resistance of the circuit will be capital r plus molar as in turn resistance is always in series.
01:16
So current in the circuit is.
01:21
I is equal to net emf divided by net resistance which becomes 24 divided by r plus 3.
01:36
So the power consumed across r will be given by the expression i squared into r.
01:56
Here this power is 21 watt for i this is 24 by r plus 3 to the whole square into r.
02:15
If we open the bracket, this is square of 24 in the denominator the identity a plus b to the whole square, it becomes r square plus 6 r plus 9 into r.
02:29
So it becomes square of 24.
02:38
Here comes out to be 5776 r divided by r squared plus 6 r plus 9.
02:51
Making a cross multiplication, it becomes 21 r square plus 126 r plus 189 is equal to 576...