Determine the current and power expended in each resistor in

Determine the current and power expended in each resistor in the bank shown below. A section of wire of total length 200.0 m is measured to have a current density magnitude of 0.538 MA/m^2 when connected to a battery providing a potential difference of 3.00 V. What is the wire material made of?

Solution

R2,R3,R4 are in series.Resultant of these three is R = R2+R3+R4

           R = 3 + 4+ 8 = 15 ohm

R6,R7 are in series.Resultant of these two is R \' = R6+R7

          R \' = 1+ 6 = 7 ohm

R ,R \' ,R5 are in parallel.

Let the resulatnt be R \" then

(1/R \") =(1/R) +(1/R \') +( 1/R5)

          =(1/15)+(1/7)+(1/2)

          = 0.7095

    R \" = 1.409 ohm

R \" ,R1 ,R8 are in series.

Equivalent resistance Req = R\" +R1+R8

                                      = 1.409 +5+7

                                      = 13.409 ohm

Potential difference V = 100 volt

Current inn the cirucit i = V/Req

                                  = 100 / 13.409

                                  = 7.457 A

Current through R1 is i = i

So, power in R1 is P1 = i ² R1

                                = 7.457 ²(5)

                                = 278 W

Ratio of currents in R,R5,R \' are = (1/R) :(1/R5):(1/R\')

                                               = (1/15):(1/2):(1/7)

                                              = (210/15) :(210/2):(210/7)

                                              = 14 : 105 : 30

Current through R is i \' = i [ 14/(14+105+30)]

                                 = 7.457 [14/149]

                                 = 0.7 A

Current through R5 is i \'\' = i [ 105/(14+105+30)]

                                 = 7.457 [105/149]

                                 = 5.254 A

Current through R \' is i \'\" = i [ 30/(14+105+30)]

                                 = 7.457 [30/149]

                                 = 1.501 A

Power through R2 is P2 = i \' ² R2

                                   = 0.7 ²(3)

                                   = 1.47 W

Power through R3 is P3 = i \' ² R3

                                   = 0.7 ²(4)

                                   = 1.96 W

Power through R4 is P4 = i \' ² R4

                                   = 0.7 ²(8)

                                   = 3.92 W

Power through R5 is P5 = i \'\' ² R5

                                   = 5.254 ²(2)

                                   = 55.2 W

Power through R6 is P6 = i \"\' ² R6

                                   = 1.501 ²(1)

                                   = 2.253 W

Power through R7 is P7 = i \"\' ² R7

                                   = 1.501 ²(6)

                                   = 13.51 W

Power through R8 is P8 = i ² R8

                                   = 7.457 ²(7)

                                   = 389.24 W