0:00
Hi.
00:01
The given problem here, this is the circuit diagram which is having a resistance of 10 oom joined in series with another resistance of 20 oom and a single battery having the emf of 15 volt.
00:27
Now, the net current passing through this circuit will be given by oms law as net emf divided by net resistance.
00:37
So it becomes 15 by 10 plus 20 oom.
00:45
So this is 15 volt divided by 30 om or we can say this current here comes out to be 0 .5 ampere.
00:59
Now due to this current, the potential drop taking place across the first resistor will be using woms law again that is given by v1 equals to i into r1 means this is 0 .5 into 10 means this is 5 volt and potential drop across the second resistance again using homs law i into r2 so this is 0 .5 into 20 means this is 10 volt so these are the two answers for the first part of the problem now we have to plot a curve for the potential drop and the position along this circuit starting from this lower most point at the lower left, where the potential is supposed to be zero.
01:57
So starting from this point, if we plot a curve in the second part of this problem, part b.
02:07
Initially, there was no potential at the chosen point after which the potential suddenly increases due to the applied battery and it increases up to the 15 volt.
02:22
Here this is potential v and here this is the position along the circuit...