Question about circuits with transistors (current, voltage)

[Mohamed Abdul]

Homework Statement

1. In Figure A-1 of Experiment 18, assume that Vdc = 8Volts, R1 = 1kOhms, R3 = 7kOhms and there is no R2 resistor. Determine the current in R1 when the transistor is fully on and carrying current. Express your answer in mA.
2. In the circuit of Figure A-1, Experiment 18, assume that Vdc = 19Volts, R1 = 4kOhms, R2 = 26kOhms and R3 = 5kOhms. When the transistor is off, determine the output voltage (also the collector voltage) in Volts.

The figure in question is here:

Homework Equations

When the transistor of a circuit is off, current is 0 amps.
When the transistor is on, current through R1 and R2 is the same (not sure but I think this is true since the circuit would short circuit).

The Attempt at a Solution

For the first one, I assume that the current through R1 would just be the Voltage/1000 = 8mA.
For the second one, I think the output voltage would just be calculated as Vin*(10^6+10^3)/(10^6) = 5.005 volts. I think that answer would make sense since the transistor is off, so the circuit wouldn't short circuit and we could take R2 into account.

 

[donpacino]

current through R1 and R2 is the same

didnt you say there was no R2. That would make the resistance inf

Also when the transistor is off, the current will be 0.
When the tranistor is on, you need to show more detail.

 

[Mohamed Abdul]

didnt you say there was no R2. That would make the resistance inf

Also when the transistor is off, the current will be 0.
When the tranistor is on, you need to show more detail.

What would be the process to find the current in that scenario, then. I assumed that since it was in the resistor before the transistor I could just use Vc/R1.

 

[donpacino]

so in problem 1, R2=inf.
You should be able to solve for Ib using Vbe, Vin, and R3.
Then, Ie=Ib*(beta+1)
You know Ib+Ic=Ie
solve for Ic

for problem 2 its actually a bit easier

 

[rude man]

For the second one, I think the output voltage would just be calculated as Vin*(10^6+10^3)/(10^6) = 5.005 volts. I think that answer would make sense since the transistor is off, so the circuit wouldn't short circuit and we could take R2 into account.

where do you get 10^6? look at what R2 is given again. Then use the correct voltage divider expression.