00:01
For this question, question 69, we want to first find what is the overall equivalent resistance of the circuit and from there find the current from the battery.
00:18
Now to do that, we will need to simplify the resistors that are in parallel.
00:25
What we have to do is to add them up together.
00:27
So for the r1 and r2, r1, r2 portion, we can simplify them by adding them up together in resprocals.
00:39
So 1 over 9 plus 1 over 18, inverse to find the effective resistance across this particular part of the circuit.
00:56
We get 6 oms.
00:58
We do the same for the r3r4 portion.
01:05
So this 1 over 10 plus 1 over 10, we get 5 oms.
01:14
After which we can find all the equivalent resistance by summing them together, since they are all now in series.
01:23
So plus r5.
01:25
So we get 6 oms plus 5 plus 8 oms from the r5 plus 1 ome from the internal resistance of the battery.
01:36
So 6 plus 5 plus 8.
01:38
This one, we get 20 oms.
01:41
This is the equivalent resistance of the entire circuit.
01:47
And from there we can find out what is the current.
01:54
By taking i equals to v over r.
01:59
V, which is the emf of the battery, 30 volts, divided by the total resistance of the circuit, which is 20 oms gets 1 .5 ampers.
02:19
Now the next part we want to do is to find the potential drop across each resistor and to do that we will need to know what is the in fact the across all of these resistors we can use v equals to ir to find the potential drop.
02:55
For the parallel circuits, we can use the effective resistance to find what is the potential drop because they both share the same potential drop, right, since they are connected in parallel.
03:11
So let us start off with the easiest one, which is the potential drop across r5...