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@TuringTest @Preetha @Omniscience @amistre64 @Algebraic!
ok lets try this.. can you find the total resistance of the circuit?
1/4 + 1/6 + 2 = total resistance
as a matter of fact no.. 1/r = 1/r1+1/r2
can u show the working
first just try to find the resistance in parallel then add it in series one so u dont make mistake.. and ok let me
first just find the parallel part of resistance.. 1/r = 1/6+1/4 1/r = 5/12 agreed?
ok total resistance = 12/5 + 2
yes.. that's it.. so you have R now.. you also have V given so appply V=IR find I from here it will be your A1
ok wt will A2 nd A3 be?
ok.. now if you know.. when a resitor in series there is a drop in voltage.. and in parallel all resistors have same voltage.
so find the drop on 3ohm resistor which is in series by V=IR
i dont get that part. now the current in a1 = 2.72 amps, so the voltage is 12v?
the volatge of whole circuit is 12V .. now you have to find the drop in voltage across just 3ohm resistor.. means a voltage is decreased at the 3 ohm resitor.. you have to find how much 3ohm took and what is remaining left.. like this .. if A=2.72 and R=3 V=2.72*3 = ?
oh okay...so voltage = 8.18 V, then what do i do
ok means from 12V.. 3ohm took 8.18V.. how much voltage is left.?
ok so now 3.81 V will move on.. now the part is parallel resistors .. when resistors are in paralle they both have same voltage... so 6ohm and 4ohm resistor have same voltage. now again you can find I by V=IR
means both the resistors hve 3.81 V
because they are in parallel..
okay thanks...great help du knw anyother easy method of solving this
lol .. its pretty easy.. you just have to do more practise.. it just take 1 min for me to solve.. maybe you can ask someone else for other method.. but this is the only one i know for now.
i am not actually great at these, but i am worried you have some mistakes. The total resistance is 12/5 + 3 = 5.4 ohms. Then the current A1 = 2.22 A. Plus now, it is easy to split the current between the 4 and 6 ohm resistors. the 4 ohm gets (4/10) * 2.22 , the 6 gets (6/10) * 2.22. Does this help/do you agree?
i am sorry.. i havent checked the calculations.. i thought he might have done right.. i was just trying to convey the concept.. and ya i think i agree.