SUM1 pl. explain this 2me.....................

[4][4] k [4][4]

Oh wow!

tht wud b wonderful

[1][1]

lukin forward to it.......

i m just pasting the circuit for each step

nahi step 1 main R3 use hoga find net current 5 and 4 ohms in parallel so net resistance 20/9 hoga net i=27/10 current in req branch dat is 2 ohm and 3 ohm wala branch will be 6/5 A

in step 1 we will certainly use R3 but in next 2 steps u replace R1 by a wire with 0 resistance so whole of current will pass through it and R3 will be useless thats why i didnt use it in step 2 and 3

[4][4][4][4]

answer nahiin match hora kya? tera method toh sahiin hain. use the fact that it offers infinite resistance in steady state

k we use superposition principle

another thing we hav to find V btw E2 and end of R2

ie ---------||-----------/\/\/\/\----------- ----------------> (arrow shows the direction of current)

||=E2 /\//\/\/\=R2

now step 1:

replace E2 and E3 by wires and find the current in above part of circuit shown with only E1

we will get a current 6/5 A parallel to the given directional arrow ( which i showed above)

step 2

similarly replace E1 and E2 and calculate current in req part of circuit

it comes out to be 2/5 A opposite to the arrow i showed

similarly step 3

using only E3 we get 3/5 A in opp dir to ma arrow

so net current in our required part of circuit is 1/5A parallel to ma arrow

now find the required potential drop V=12/5 V

so energy = 5*12*12/2*5*5=144/10=14.4 units

[4][4][4]

YAH VISH

u r crrct bro! ans givn is C

Can u pl. post ur workin in a lil mor detailed way

i did it by superposition and of coure R3 has to be considered for it

k i will get the ans and tel

l i m taking it 4 ohm

lagta aisa hi he.... the symbol is lil wierd but I m not sure if its ohm or sumthing else..... par in da diag resistor only has bn shon

R3 kitna hai 4 ohm ????

DING DONG!!!

KOI HAI?

NO.
ACTUALLY, my methd dusnt match wid da abov mentioned methd..............
its a que frm Tiit FT1

but y do v not consider R3?????????????/