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Explain the meaning of the underlined phrase and calculate the transmitting frequency.

Physics

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Here are the solutions for the questions from Section III, Option 3:

OPTION 3: ELECTRONICS

1. (a) (i) The underlined phrase "a bandset clearly picks the signals from the radio station" refers to the tuning or resonance phenomenon in a radio receiver circuit. A tuning circuit is designed to resonate at a specific frequency, allowing it to select and amplify signals at that frequency while rejecting others.

1. (a) (ii)
Assumption: The circuit in Figure 4 is an LC resonant circuit, and the inductor has a value of L=12 mHL = 12 \text{ mH} (based on typical values for such problems).

  1. Convert given values to SI units.

    Inductance L=12 mH=12×10−3 HL = 12 \text{ mH} = 12 \times 10^{-3} \text{ H}
    Capacitance C=2.7 pF=2.7×10−12 FC = 2.7 \text{ pF} = 2.7 \times 10^{-12} \text{ F}

  2. Use the formula for resonant frequency ff.

    f=12πLCf = \frac{1}{2\pi\sqrt{LC}}

  3. Substitute the values and calculate ff.

    f=12π(12×10−3 H)×(2.7×10−12 F)f = \frac{1}{2\pi\sqrt{(12 \times 10^{-3} \text{ H}) \times (2.7 \times 10^{-12} \text{ F})}} f=12π32.4×10−15 s2f = \frac{1}{2\pi\sqrt{32.4 \times 10^{-15} \text{ s}^2}} f=12π3.24×10−14 s2f = \frac{1}{2\pi\sqrt{3.24 \times 10^{-14} \text{ s}^2}} f=12π×1.8×10−7 sf = \frac{1}{2\pi \times 1.8 \times 10^{-7} \text{ s}} f=11.13097×10−6 sf = \frac{1}{1.13097 \times 10^{-6} \text{ s}} f≈884200 Hzf \approx 884200 \text{ Hz} f≈884.2 kHzf \approx 884.2 \text{ kHz}

    The transmitting frequency is 884 kHz\boxed{\text{884 kHz}}.

    1. (b) (i) The underlined phrase "d.c. gain of 100" means that for every unit of direct current (d.c.) flowing into the base of the transistor, 100 units of d.c. will flow through the collector. It is the ratio of the collector current to the base current (βdc=ICIB\beta_{dc} = \frac{I_C}{I_B}).

    1. (b) (ii)

  4. Identify the given values.

    d.c. gain βdc=100\beta_{dc} = 100
    Base current IB=22.5μAI_B = 22.5 \mu A

  5. Calculate the collector current ICI_C.

    IC=βdc×IBI_C = \beta_{dc} \times I_B IC=100×22.5μA=2250μAI_C = 100 \times 22.5 \mu A = 2250 \mu A IC=2.25 mAI_C = 2.25 \text{ mA}

  6. Calculate the emitter current IEI_E.

    The emitter current is the sum of the base and collector currents.

    IE=IB+ICI_E = I_B + I_C IE=22.5μA+2250μA=2272.5μAI_E = 22.5 \mu A + 2250 \mu A = 2272.5 \mu A IE=2.2725 mAI_E = 2.2725 \text{ mA}

    The collector current is 2.25 mA\boxed{\text{2.25 mA}} and the emitter current is 2.27 mA\boxed{\text{2.27 mA}}.

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