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