Verified Solution • 2025–26 Session moderate

QNEET-2018-Ph-897: NEET 2018 Physics - Class 12 NEET-2018-Ph-897 Solution: In the circuit shown in the figure the input voltage \(V_i\) is …

In the circuit shown in the figure the input voltage ViV_i is 20 V VBE=0V_{BE} = 0 and VCE=0V_{CE} = 0. The values of IBICI_B I_C and β\beta are given by IB=40μAIC=10 mAβ=250I_B = 40\mu A I_C = 10\ \text{mA} \beta = 250 IB=25μAIC=5 mAβ=200I_B = 25\mu A I_C = 5\ \text{mA} \beta = 200 IB=20μAIC=5 mAβ=250I_B = 20\mu A I_C = 5\ \text{mA} \beta = 250 IB=40μAIC=5 mAβ=125I_B = 40\mu A I_C = 5\ \text{mA} \beta = 125

Multiple Choice Options

  • A I B = 40 μ\mu _ A, I C = 10 mA, β\beta _ = 250
  • B I B = 25 μ\mu _ A, I C = 5 mA, β\beta _ = 200
  • C I B = 20 μ\mu _ A, I C = 5 mA, β\beta _ = 250
  • D I B = 40 μ\mu _ A, I C = 5 mA, β\beta _ = 125

Step-by-Step Verified Solution

From question, V BE = 0, V i = 20 V V CE = 0 V b = 0 (earthed) We need to find values of base and collector currents I B and I C and amplification factor _ , so V CC = I C R C + V CE I C = V CC � V CE /R C = 20 V � 0 V/4000 = 5 mA Base current I B = V i /R B = 20 500 _ 10 3 = 40 _A Amplification factor _ = I C I B = 25 _ 10 _ 3 40 _ 10 _ 6 = 125

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