In the circuit shown in the figure, the input voltage V _i is 20 V, V _BE = 0 and V _CE = 0. The values of I _B , I _C and β are given by 1. I_B = 40 μ A, I_C = 10 mA, β = 250 2. I_B = 20 μ A, I_C = 5 mA, β = 250 3. I_B = 25 μ A, I_C = 5 mA, β = 200 4. I_B = 40 μ A, I_C = 5 mA, β = 125

Answer: 4. I_B = 40 μ A, I_C = 5 mA, β = 125 From the figure: R_B=500\ kΩ , R_C=4\ kΩ , supply 20 V, V_i=20 V, with V_BE=0 and V_CE=0 . I_B=dfracV_i-V_BER_B=20/500×10³=40\ μA . V_CE=0 means the whole 20 V drops across R_C : I_C=20/4×10³=5\ mA . β=I_C/I_B=frac5×10⁻³40×10⁻⁶=125

Topic: Semiconductors· Electronic Devices

In the circuit shown in the figure, the input voltage V_i is 20 V, V_BE = 0 and V_CE = 0. The values of I_B, I_C and β are given by

In the circuit shown in the figure, the input voltage V$_i$ is 20 V, V$_{BE}$ = 0 and V$_{CE}$ = 0. The values of I$_B$, I$_C$ and $\beta$ are given by

transistor circuit with RB, RC, 20 V

1. $I_B$ = 40 $\mu$A, $I_C$ = 10 mA, $\beta$ = 250

2. $I_B$ = 20 $\mu$A, $I_C$ = 5 mA, $\beta$ = 250

3. $I_B$ = 25 $\mu$A, $I_C$ = 5 mA, $\beta$ = 200

4. $I_B$ = 40 $\mu$A, $I_C$ = 5 mA, $\beta$ = 125

Doubtly Asked 30 Oct 2026 NEET UG 2018 neet Save WhatsApp Report

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Answer: 4. $I_B$ = 40 $\mu$A, $I_C$ = 5 mA, $\beta$ = 125

From the figure: $R_B=500\ \mathrm{k\Omega}$, $R_C=4\ \mathrm{k\Omega}$, supply 20 V, $V_i=20$ V, with $V_{BE}=0$ and $V_{CE}=0$.

$I_B=\dfrac{V_i-V_{BE}}{R_B}=\dfrac{20}{500\times10^3}=40\ \mu\mathrm{A}$.

$V_{CE}=0$ means the whole 20 V drops across $R_C$: $I_C=\dfrac{20}{4\times10^3}=5\ \mathrm{mA}$.

$$\beta=\frac{I_C}{I_B}=\frac{5\times10^{-3}}{40\times10^{-6}}=125$$

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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In the circuit shown in the figure, the input voltage V_i is 20 V, V_BE = 0 and V_CE = 0. The values of I_B, I_C and β are given by

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