In the following circuit R10813 Ohm R22 13 Ohm R33 13 Ohm R4

In the following circuit, R1=0.8*13 Ohm, R2=2* 13 Ohm, R3=3* 13 Ohm, R4=2* 13 Ohm, and E= 36.2 Volt. Calculate the power Loss in R2 resistor? [Picture File-Ohm-3.jpg]

Solution

R1 = 0.8(13 ohm)

     = 10.4 ohm

R2 = 2(13 ohm)

    = 26 ohm

R3 = 3(13 ohm) = 39 ohm

R4 = 2(13 ohm) = 26 ohm

E = 36.2 volt

R2 andR3 are in parallel.

Resultant of these two is R then (1/R) = (1/r2)+(1/R3)

(1/R) = (1/26)+(1/39)

          = 0.0641

     R = 15.6 ohm

R1,R,R4 are in series.

Equivalent resistance Req = R1+R+R4

                                      = 10.4 ohm + 15.6 ohm +26 ohm

                                      = 52 ohm

Current in the circuit i = E / Req

                                = 36.2 / 52

                                = 0.6961 A

Current in R2 is i \' = i[R3/(R2+R3)]

                          = 0.6961 [ 39 /(26+39)]

                          = 0.4176 A

Power loss in R2 is P = i\' ² R2

                                 = (0.4176) ² (26)

                                 = 4.536 watt