00:01
High in the given problem here this is the circuit diagram having first of all one of the resistor having a value of 3 .00 oam then another resistor here having a resistance of 8 .00 oom then the bottom part having none of the resistance then in the right arm, one resistance having 1 .0 own value and a battery having the polarity as shown here, positive, negative, and having a potential difference of 12 .0 volt, then a middle arm having two resistances, 5 .00 oom, and 1 .00 oom.
00:59
And then a battery having a value of potential difference as 4 .00 volt and polarity as shown here in the figure now in the first part of the problem we have to find we have to show how to add just enough amometers to measure every different current so we know amometer should be connected in series to measure the the current because we know current remains same in series.
01:34
Hence, as there are three branches, there are three arms of the circuit, so we should join just three amometers.
01:43
The first emmeter we should join here.
01:48
This is a1.
01:51
We can say it.
01:53
Its positive terminal should be towards the positive terminal of the cell and negative terminal should be towards the negative terminal of the cell.
02:01
Then we join the second ammeter here.
02:07
We can join it here also.
02:09
As the current remains same in series.
02:13
This is a2.
02:15
Its negative terminal is towards the negative of the cell and positive on the other side...