In the circuit shown here, the transistor used has a current gain β = 100. What should be the base resistor RB so that VCE = 5 V (neglect VBE)?

Question

In the circuit shown here, the transistor used has a current gain β = 100. What should be the base resistor RB so that VCE = 5 V (neglect VBE)? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/92979ad3fb4d7c8c3df2a45f485b7f25-.png" /> (A) 200 × 103 Ω (B) 1 × 106 Ω (C) 500 Ω (D) 2 × 103 Ω

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9c38edc77b2b98e6bce62d84fdbd8bb6-.png" /> Applying Kirchhoff's law to mesh ADCEFA VCC =10 = IC × 103 +VCE 10 = IC × 103 + 5 IC = 5 × 10-3 A ...(i) Applying Kirchhoff's law to mesh AGBEFA IB × RB + VBE = VCC = 10 V IB × RB + 0 = 10 ... (ii) 5 × 10-5 × RB = 10 RB = 2 × 105 Ω = 200 × 103 Ω

Q. In the circuit shown here, the transistor used has a current gain $β = 100$ . What should be the base resistor $R_{B}$ so that $V_{CE} = 5 V$ (neglect $V_{BE}$ )?

$200 \times 1 0^{3} Ω$

$1 \times 1 0^{6} Ω$

$500 Ω$

$2 \times 1 0^{3} Ω$

Q. In the circuit shown here, the transistor used has a current gain β=100. What should be the base resistor RB​ so that VCE​=5V (neglect VBE​)?

200×103Ω

1×106Ω

500Ω

2×103Ω

Applying Kirchhoff's law to mesh ADCEFA VCC​=10=IC​×103+VCE​ 10=IC​×103+5 IC​=5×10−3A...(i) Applying Kirchhoff's law to mesh AGBEFA IB​×RB​+VBE​=VCC​=10V IB​×RB​+0=10...(ii) 5×10−5×RB​=10 RB​=2×105Ω =200×103Ω

Q. In the circuit shown here, the transistor used has a current gain $β = 100$ . What should be the base resistor $R_{B}$ so that $V_{CE} = 5 V$ (neglect $V_{BE}$ )?

$200 \times 1 0^{3} Ω$

$1 \times 1 0^{6} Ω$

$500 Ω$

$2 \times 1 0^{3} Ω$