Questions 7 and 8: In the circuit shown below, the current...

Transcript

00:01 Hello everyone so let's start with the question as you can see in the diagram there is a connection of resistence and voltage and ammeter so we have to find the current through 40 om so here we have to find what the current through this resistor and the second we have to find the resistor r that is this value okay, so let's start with the solution so here first we have to understand some points okay so whenever the current is flowing from one point to another point the current current value will remain same that is i will write it down that is current that on that flow through any point any point or nodal will be equal to the current that leaves normal okay so here as you can see in the diagram we have to find what we have to find current in register 40 om okay so here when the current passes through this ammeter and it comes towards the 40 om the current will remain same as per the point okay so here it will not there will be no any changes so so our answer will be what that is as you can see that in ammeter it's it shows 20 ome okay as the in a question it is given that in ammeter the value shows what 20 millie ampere and here the volt meters have that is 1 volt okay so here our the value of current is 20mphal it means it detect 20m ampere after passing through 40 om it will remain same so as per the point when the current flow through any point or nodal will be equal to the current that leads through point okay so our answer will be what our answer will be for the current for the resistor 40 om our current will be our current our current will be that is current i will be 20 mili amp so this is the answer of our first question let's select the option that is our d option will be correct now in second question now in second question we have to find what resistance are okay we have to find the resistant value here so let's start the solution as i if i draw it if i draw a diagram here okay so here as that this and this are parallel to each other okay if this 15 om and 30 oam are parallel to each other 15 om and 30 om are parallel to each other so each other so if our r net will be what that is 15 multiply by 30 that is a we are using parallel that is one upon r equal to one upon r1 plus 1 upon r2 okay after solving this that is 15 plus 30 after solving this our answer will be what 10 oh okay now this 10 ome up to means if i redraw this if i redraw this that is volt here 40 om this is our m meter this is our 10 and this is our 20 o this is our r this our voltage okay and this is our 20 oom and this is and this is 3 .3 volt okay so as you see in the diagram this 40 and 10 are in series so our r net will be that is our r dash net will be what 10 plus 50 sorry 10 plus 40 10 plus 40 10 plus 40 that is 50 oh okay so this is the value of our like this series so if i use uh kirtzhov voltage low so according to the kirchof voltage low in short if i say kvl rule so as for the kval rule if there are two two cycles as you can see there are two registers box so we will use our kirchov voltage low and if we have to draw like we have to go through in clockwise and anti -clockwise whatever the direction we will draw suppose if i go in clockwise direction in this way that is if this is our first and this is our second, so as per the kirchow law in i am applying kirchow flow in loop 1 so our value will be what that is minus 50 multiply by i plus i are multiply by multiply by capital are equal to 0 means in this as you see if if after solving this this is our 50 so if i apply this then the current will be flow in this direction that is 50 oh and then i that is in i and plus in this there will be current i r and r okay so our ir will be what ir multiplied by r equal to 50 i okay here the i is what this is the current that is flowing ammeter that is the current that is flowing ammeter so our i is what 20 millie ampere okay so i are multiply by r so 50 20 millie ampire 20 10 to minus 3 ampere after solving this our answer will be what 1 ampere okay so ir multiply by r this is our first now i can write that is i i can also write that is our irr equal to 1 upon r okay now if i apply kirch of law in second loop applying kvall on loop 2 so what it will be and also we know that also we know that that that is i0 is equal to i plus ir okay what it is mean means whenever there is a junction suppose say there is junction if suppose the current is flowing that is i not it divide into two part that is i and i so as you can see there is a current that is flowing from here that is i not when it goes to this point this is a junction so it will divide into two part that is i and that is ir okay so this is the addition of this i note is the addition of i plus ir okay now so in applying kvial rule so answer will be what 3 minus 20 i0 minus 1 minus 20 i not equal to 0 therefore 3 minus 20 anode and 20 anode what for t i not minus 1 equal to 0 3 minus 1 that is 2 equal to 40 i0 therefore our i not is what 1 upon 20 now as you can see that is as you can see that our i not is what i plus i r so i plus i are equal to 1 upon 20 okay here our i is what 20 milli amp that is given in the question 20 10 multiply by 10 to minus 3 amp plus ir is what 1 upon r okay as we have find that is here i are equal to 1 upon r so equal to 20 1 upon 20 therefore are taking lcm that is r that is r that is r multiply by 20 and is 2 minus 3 plus 1 equal to r will go in this direction 20 and 20 will go in this direction that is in this direction so it will be what 400 multiply by r multiply by 10 is to minus 3 plus 1 equal to r okay after solving this equation after solving this is where r will be what 20 multiply by 10 divided by 6 and after solving this our r value will be what 33 .33 that is approximate 33 so, so this is our final answer.

10:07 Now select the options.

10:09 So our option will be what for the that is d option...

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