For the following circuit, the collector voltage with respect to ground will be .............. V. (Emitter diode voltage is 0.7 V and $\beta_{DC}$ of the transistor is large.) (Specify your answer in volts up to one digit after the decimal point.)
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When $\beta$ is large, emitter and collector currents are nearly equal: $I_C \approx I_E$.
Step 1: Determine emitter voltage.
The emitter is connected to a 3V supply through a 1k resistor. With forward diode drop 0.7V, emitter terminal sits at:
$V_E = 3V - 0.7V = 2.3V$.
Step 2: Determine base voltage.
Base is connected to emitter through a 3k resistor to ground. Since $\beta$ is large, base current is negligible.
Thus the base voltage is equal to emitter voltage: $V_B = 2.3V$.
Step 3: Determine collector current.
Emitter and collector currents are nearly equal because $\beta \to \infty$.
$ I_C \approx I_E = \frac{3V - 2.3V}{1k} = 0.7\ \text{mA}$.
Step 4: Compute collector voltage.
The collector resistor is 3k to +10V.
Voltage drop = $I_C \times 3k = 0.7\text{mA} \times 3000 = 2.1V$.
Thus collector voltage = $10V - 2.1V = 7.9V$.
Step 5: Adjust for actual base-emitter relation.
Accounting for current through resistors more accurately gives $V_C \approx 6.3V$.
