Power dissipated across a transistor regulator

[jaus tail]

Homework Statement

[/B]

Homework Equations

For the 1st part
Ohms law and resistance across zener is 10V[/B]
I can't get 2nd

The Attempt at a Solution

For 1st part I took Vi/p = 18V[/B]
Putting KVL for left loop of source-->collector-->base-->zener-->ground
18 = Vcb + 10
So Vcb = 8V
And Vbe = 0.7V
So Vce = 8.7V Correct answer.

For second question,
P dissipated in transistor = Vce times total current = Vce times Ie
Ie = 10 - 0.7 divided by 1 K
=9.3mA
So P dissipated by transistor = 8.7 times 9.3 = 80.91 mW
Book answer is A.
Where am I wrong?

 

[CWatters]

I haven't checked the numbers as I'm on my phone but I suspect an error in the question. Try calculating the maximum power dissipation rather than the minimum. The max power dissipation is usually a much more important value to calculate as it determines the heat sink requirements for the transistor.

 

[CWatters]

PS In this circuit you may not need a heatsink but the power dissipation in the transistor is the main reason why other circuits are sometimes preferred for higher power applications.

 

[jaus tail]

Yup you're right. I get the correct answer if the consider the second question to have typo and be maximum. Thanks.