Question

in the circuit shown below the diode has a forward voltage drop of 0.5 v. what is the power dissipated?

Answer 1

In the circuit shown below the diode has a forward voltage drop of 0.5 v. what is the power dissipated?

Let say R is resistance of Diode

then R+1 parallel to 3

= 1/(1/(R+1) + 1/3)

= 3(R+1)/(R+4)

total resistance of circuit

= 3(R+1)/(R+4) + 1

=(4R + 7)/(R+4)

I = 6.5/((4R + 7)/(R+4))

let say I1 current through

diode circuit

I1R = 0.5V

I1(R+1) = I2(3)

I1(R+1) = (I -I1)3

I1R + I1 = 3I -3I1

I1R + 4I1 = 3I

I1R + 4I1 = 3*6.5/((4R + 7)/(R+4))

(0.5 + 4I1)(4R+7) = 19.5(R+4)

2R + 16I1R + 3.5 + 28I1 = 19.5R + 78

2R + 8 + 3.5 + 28I1 = 19.5R + 78

28I1 - 17.5R - 66.5 = 0

14/R - 17.5R -66.5 = 0

28/R - 35R -133 = 0

-35R^2 -133R + 32 = 0

35R^2 + 133R -28= 0

35R^2 +140R -7R -28 = 0

35R(R+4) -7(R+4)= 0

R = 7/35 = 1/5 = 0.2 ohm

V^2/R = 0.5 * 0.5/0.2

= 1.25 w