Find the value of I_B if V_{BE} = 0.3,V in th | vedclass

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(A) For the input loop,applying Kirchhoff's Voltage Law $(KVL)$:$6 - (10 	imes 10^3) I_B - V_{BE} - (1 	imes 10^3) I_E = 0$Given $\beta = 100$,we kn

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$51$

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(A) For the input loop,applying Kirchhoff's Voltage Law $(KVL)$:$6 - (10 	imes 10^3) I_B - V_{BE} - (1 	imes 10^3) I_E = 0$Given $\beta = 100$,we know $I_E = I_B + I_C = I_B + \beta I_B = (1 + \beta) I_B = 101 I_B$.Substituting the values:$6 - 10000 I_B - 0.3 - 1000(101 I_B) = 0$$5.7 - 10000 I_B - 101000 I_B = 0$$5.7 = 111000 I_B$$I_B = \frac{5.7}{111000} \approx 51.35 	imes 10^{-6} A = 51.35\,\mu A$.Rounding to the nearest given option,the value is $51\,\mu A$.

Find the value of $I_B$ if $V_{BE} = 0.3\,V$ in the given circuit $(\beta = 100)$. (in $,\mu A$)

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