Question

find the equivalent resistance of the following circuit .also find the current and potential at each resistor​

Answer 1

hope this helps u

Explanation:

r2,r3 , r4 are in parallel

1/rp = 1/1+1/2+1/3 lcm of 2,1,3 = 6

      =  6+3+2/6

       = 11/6

rp = 6/11= 0.5454 ohm

r1 , r5 , r4 is in series

2+6/11+2

22+6+22/11

50/11

total resistance is 50/11 = 4.5454ohm

we know that v = ir

here r = 4.54 and potential difference = 9 v

i = v/r

 = 9/4.54

  = 900/454

i=1.98 ampere

Answer 2

Answer:

1/R'=1/R2+1/R3+1/R4

1/1+1/2+1/3=6+3+2/6=11/6

R'=6/11 ohm

now,the given circuit can be redraw as shown below

now, R1 and R2 are in series combination. as, current through R1,R' and R5 is same

so, equivalent resistance of the whole circuit is

R= R1+R'+R5=2+6/11+2

22+6+22/11=50/11

now,total current flowing through the circuit

l=V/R=99/50=2A

current through R1 and R5 will be same as these are in series combination and will be equal to the total current flowing through the circuit

l=I1=I5=2A

Potential drop atR1, V1 =I1R1=2×2=4V

now, potential drop at R',V' can be calculated as

V=V1+V5+V'

9=4+4+V'

V'=1V

As R2,R2 and R4 are in parallel combination,so potential drop at all resistance will be same as 1 V

V2=V3=V4=V'=1V

current through R2,I2=V2/R2=V'/R2=1/1=1A

similarly,. I3=V3/A3=V'/R3=1/2=0.5A

and. I4=V4/R4=V'/R4=1/3=0.33A