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I have found out the output voltage to be 9V.

To find power, we need \$I_CV_{CE}\$

We can find \$I_E\$ from the output circuit.

\$ I_E=\frac{6}{24k}+\frac{9}{10} \$

But can we find \$I_C\$? Or do we have to approximate \$I_C\$ approximately equal to \$I_E\$?

Will increase of \$ V_{in}\$ by 20% anyway impact the transistor currents?

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  • \$\begingroup\$ The output current is not 9V, current is measured in amps... \$\endgroup\$ – Solar Mike Oct 28 '18 at 10:59
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    \$\begingroup\$ Since beta is not given you can assume the base current is negligible. That's the safer assumption (from a dissipation point of view anyway). Increasing the input voltage will increase the dissipation. The effect on the transistor currents will be small. \$\endgroup\$ – Spehro Pefhany Oct 28 '18 at 11:03
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For this theoretical circuit, the output voltage will be around: $$V_O = 6V \cdot (1 + \frac{12\textrm{k}\Omega}{24\textrm{k}\Omega}) = 9V$$

Hence the output current is: \$I_O = \frac{9V}{10\Omega} = 0.9A\$

The power dissipation is \$P \approx (V_{IN} - V_O) \cdot I_O = 5.4W\$

I assumed that \$I_C \approx I_E\$ because \$\frac{I_C}{I_E} = \frac{I_C}{I_B+I_C} = \frac{ \beta I_B}{I_B +\beta I_B} = \frac{ \beta I_B}{(\beta +1) I_B} =\frac{ \beta}{\beta +1} \$

So if the beta is large we can assume that \$I_C \approx I_E\$

And now you should be able to calculate the power dissipated in the transistor after voltage increases.

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