# BJT - finding Ic with two opposite voltages connected to base

Here's a question from my homework:

Find $$\ I_C \$$ and $$\ V_{CE} \$$ for the circuits below, assuming $$\ |V_{BE}| = 0.7V \$$ and $$\ \beta = 100 \$$. I understand circuit (a), but for (b), I don't understand how to incorporate the $$\ -15V \$$ branch into calculation.

Below is my attempt:

$$R_{TH} = \frac {R_1 R_2} {R_1 + R_2} = 3.197 \times 10^5 \Omega$$

$$V_{TH} = \frac {V_1 R_2 + V_2 R_1} {R_1 + R_2} = 5.408 V$$

$$\therefore I_B = \frac{V_{TH} - V_{BE}} {R_{TH}} = 1.47 \times 10^{-5} A$$

$$I_C = \beta I_B = (100) (4.69 \times 10^{-5}) = 1.47 mA$$

Edit: Thank you all so much for the helppppppppp~

• Do you know the Thevenin's theory? electronics.stackexchange.com/questions/471906/… – G36 Mar 11 at 7:40
• According to Kirchhoffs current law: Ib1=Ib+Ib2. – LvW Mar 11 at 8:38
• Disconnect the base. What is the voltage at the 2-resistor node? What is the Thevenin-equivalent resistance? Re-connect the base. – AnalogKid Mar 11 at 12:29
• -15V means that the voltage at the lower side of 1M resistor is 15V lower than the reference voltage. electronics.stackexchange.com/questions/392010/… – G36 Mar 11 at 13:06
• I see the mistake the base current is equal to $I_B = \frac{V_{TH} - V_{BE}}{R_{TH}}$ – G36 Mar 11 at 13:08