Question

Find the power dissipated in R_L

_{[10 points] Find the power dissipated in RL. RI 100 o R2 R3 Ohm}

Answer 1

In the above circuit , R2 and R3 are in series their equivallent resistance is

R' = R2 + R3 = 43 ohm + 22 ohm

                    = 65 ohm

But R1 and R' are in parallel then their equivallent resistance is

R" = ( R1 x R' ) / ( R1 + R' )

     = ( 100 ohm x 65 ohm ) / ( 100 ohm + 65 ohm )

     = 6500 ohm / 165 ohm

     = 39.39393 ohm

Now R" and R4 are in series then their equivallent is

R"' = R" + R4

       = 39.39393 ohm + 100 ohm

       = 139.39393 ohm

The output potential across R_L can be calculated as

V_out = V_s x ( R_L / R_L + R"' )

         = 10 V x ( 1000 ohm / 1000 ohm + 139.39393 ohm )   since V_s = 10 V , R_L = 1 kohm =1000 ohm

        = 10000 / 1139.39393

        = 8.7766 V

The power dissipated across R_L can be calculated by the formula P = V² /R

Here V =V_out =potential , R =R_L = load resistance

Therefore P = ( 8.7766 V )² / 1000 ohm

                    = 77.02871 / 1000

                    = 0.07702871 Volt² / ohm or watt