For the circuit shown R1415 Ohm R2245 Ohm R3125 Ohm R4485 R5

For the circuit shown, R_1=4.15 Ohm, R_2=2.45 Ohm, R_3=1.25 Ohm, R_4=4.85, R_5=1.85 Ohm, and v_ab=1.35 V. Find V.

Solution

R4 and R5 are in parallel.So, resultant resistance of these resistances R = (R4xR5)/(R4+R5)

               R = (4.85x1.85)/(4.85+1.85)

                  = 1.339 ohm

R2,R3 and R in series.So, resultant of these three resistances is R \' = R+R2+r3

           R \' = 1.339 +2.45+1.25

                = 5.039 ohm

R \' and R1 are in parallel.So, equivalent resistance of the circuit R \" = ( R \' x R1) /(R \' +R1)

                         R \" = (5.039 x 4.15)/(5.039 +4.15)

                               = 2.275 ohm

Current in the circuit i = V / R \"

                                = V /2.275

                                = 0.4394 V

Current through R1 is i \' = V/R1

                                   = V / 4.15

                                   = 0.2409 V

From Kirchoff\'s first law , current through R2 is i \" = i - i \'

      i \" = 0.4394 V - 0.2409 V

          = 0.1984 V

Potential difference across the points and b points is V _ab = i \" R2

     V _ab = 0.1984 V R2

            = 0.1984 V x 2.45

            = 0.4861 V

   1.35 = 0.4861 V

Therefore V = 1.35 / 0.4861

                   = 2.77 volt