simulate this circuit – Schematic created using CircuitLab

How do I calculate the \$I_{C}\$ from the circuit above?

I tried using \$KVL\$ using this equation:

$$-100I_{B} - 0.7 +10 = 0$$ $$I_{B} = \frac {I_{C}}{\beta}$$ $$\beta = 100$$ $$-100\frac {I_{C}}{\beta} - 0.7 +10 = 0$$ $$I_{C}= 9.3mA$$

\$I_{C}\$ should be: \$I_{C}= 1.98mA\$

I am unsure on where I went wrong.

Edit: Here is the exact question

enter image description here

  • \$\begingroup\$ Frankly the circuit, as drawn, makes little sense to me. What am I supposed to make of the two power supplies on the emitter? I dont see any voltage potential between base and emitter, tbh. V2 basically acts as a 5k resistor to ground, given the values of V2 and I2. Is the ground at the end of the emitter line at the same potential as the ground connected to the base? The base, which has no voltage applied to it, should put the xstr in cutoff. Really the whole schematic is convoluted and confusing. \$\endgroup\$ Apr 16, 2017 at 1:45
  • \$\begingroup\$ I've added another picture with the exact specs of the problem if that helps \$\endgroup\$ Apr 16, 2017 at 2:00
  • \$\begingroup\$ Does it help if you know that IE=IC+IB? And if you are unsure about this, draw a box around the transistor. Just like a node, the currents have to sum to zero. \$\endgroup\$
    – owg60
    Apr 16, 2017 at 2:15
  • \$\begingroup\$ @CogitoErgoCogitoSum Better to remove your comment. It's full of misconceptions. The schematic is perfectly valid and analyzable. This circuit can well be a part of a practical 2 transistor circuit that is real and useful. I2 can be the collector current of the other transistor.There's a valid way for Ib from gnd to -10V, the current source can be a current limiting circuit. To state something to be as 5k resistor you at first should be sure it obeys Ohms law. You use flat earther's logic. You assume V2 to be a resistor(=apply Ohm's law) and get its resistance =5k. \$\endgroup\$
    – user136077
    Apr 16, 2017 at 5:35

1 Answer 1


You can calculate back to the base current.

\$I_e = \beta * I_b +I_b\$

-> \$I_e = I_b * (\beta + 1)\$

-> \$I_b =I_e/(\beta +1) = 2/101 = .0198mA\$ (Formula 1)

\$I_c = I_e - I_b = 2 -.0198 = 1.98mA\$

You can also do it this way

\$I_c = \beta * I_b\$

Substituting (Formula 1) from above

\$I_c = I_e * \beta/(\beta + 1)\ = 2 * 100 /101 = 1.98mA\$


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