In the following common emitter circuit if -100-VCE-7V-VBE-Negligible-RC-2k- then IB-

The correct option is A 0.04mA We know that, V=V_CE+I_cR_L ⇒ 15=7+1× 2× 10' ⇒ i=4mA ∴β=i_Ci_B⇒ i_B=4100=0.04mA ...

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Question

In the following common emitter circuit if $β = 100, V_{C E} = 7 V, V_{B E} = N e g l i g i b l e, R_{C} = 2 k Ω,$ then $I_{B} = ?$

0.01 m A

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0.04 m A

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0.02 m A

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0.03 m A

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Solution

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Similar questions

β = 100, V_{C E} = 7 V, V_{B E}

R_{C} = 2 k Ω,

I_{B}

β = 100, V_{B E} = 0.7 V .

V_{C E}

β = 100,

V_{C E}

40 μ A

V_{B E}

β = 100.

V_{B E} = - 0.7 V,

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