In the following common emitter circuit- if -100- V CE -7 V - V BE - negligible- R C -2 k - then I B isA- 0-01 mAB- 0-04 mA- 0-02 mAD- 0-03 mA

The correct option is B 0.04 mAUsing kirchoffs loop law, E = V_CE + I_C R_C I_C = E V_CER_C = (15 7) V2 k Ω = 4 mA β = I_CI_B ⇒ I_B = I_Cβ = 4 mA100 = 0.0 ...

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Standard XII

Physics

Simple Circuit

In the follow...

Question

In the following common emitter circuit, if $β = 100, V_{C E} = 7 V, V_{B E}$ = negligible, $R_{C} = 2 k Ω,$ then $I_{B}$ is

0.01 mA

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0.04 mA

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0.02 mA

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0.03 mA

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Solution

The correct option is B $0.04 mA$
Using kirchoffs loop law,
$E = V_{C E} + I_{C} R_{C}$
$I_{C} = \frac{E - V_{C E}}{R_{C}} = \frac{(15 - 7) V}{2 k Ω} = 4 mA$
$β = \frac{I_{C}}{I_{B}} \Rightarrow I_{B} = \frac{I_{C}}{β} = \frac{4 mA}{100} = 0.04 mA$

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In the following common emitter circuit, if β=100,VCE=7 V,VBE = negligible, RC=2 kΩ, then IB is

The correct option is B 0.04 mAUsing kirchoffs loop law, E=VCE+ICRC IC=E−VCERC=(15−7) V2 kΩ=4 mA β=ICIB⇒IB=ICβ=4 mA100=0.04 mA

In the following common emitter circuit, if $β = 100, V_{C E} = 7 V, V_{B E}$ = negligible, $R_{C} = 2 k Ω,$ then $I_{B}$ is

The correct option is B $0.04 mA$
Using kirchoffs loop law,
$E = V_{C E} + I_{C} R_{C}$
$I_{C} = \frac{E - V_{C E}}{R_{C}} = \frac{(15 - 7) V}{2 k Ω} = 4 mA$
$β = \frac{I_{C}}{I_{B}} \Rightarrow I_{B} = \frac{I_{C}}{β} = \frac{4 mA}{100} = 0.04 mA$