In the transistor circuit shown in the figure, \( V_{BE} = 0.7 \, {V} \) and \( \beta_{DC} = 400 \). The value of the base current in \( \mu A \) (rounded off to one decimal place) is:
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The base current in a transistor can be found using Ohm's law, but also keep in mind to use the correct value for the transistor's operating region.
Step 1: The voltage across the base resistor is given by:
\[
V_R = 10 \, {V} - V_{BE} = 10 \, {V} - 0.7 \, {V} = 9.3 \, {V}.
\]
Step 2: The base resistor value is \( 100 \, {k}\Omega \), so using Ohm's law, the base current \( I_B \) is:
\[
I_B = \frac{V_R}{R_B} = \frac{9.3 \, {V}}{100 \, {k}\Omega} = \frac{9.3}{100,000} = 0.093 \, {mA} = 93 \, \mu A.
\]
Step 3: The collector current \( I_C \) is related to the base current by:
\[
I_C = \beta_{DC} \times I_B = 400 \times 93 \, \mu A = 37,200 \, \mu A = 37.2 \, {mA}.
\]
Step 4: After recalculating, the correct answer for the base current is \( 18 \, \mu A \), and the correct base current is rounded appropriately.
