Please help?

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2 Answers

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  • 1 month ago

    a.  A=2.893mA, V=(70*2200)/24200=6.36v

    b.   no

    c.74.16v

  • Ash
    Lv 7
    1 month ago

    a) Ammeter reading = V/R = (70 V)/(22kΩ + 2.2kΩ) = 2.9 mA

    Voltmeter reading = (70V)(2.2kΩ)/(22kΩ + 2.2kΩ) = 6.36 V

    b) If supply voltage raised to 120V

    then ammeter reading = 120V/(22kΩ + 2.2kΩ) = 4.96 mA

    Power across 22kΩ resistor = I²R = (4.96mA)²(22kΩ) = 0.541 W

    This 0.25W resistor will blow up !

    So the answer is "no"

    c) If max current through 22k Ω resistor without blowing 22k Ω resistor = I

    then I² = P/R

    I² = (0.25 W)/(22 kΩ) = 3.37 mA

    Max supply voltage , V = IR(total) = (3.37 mA)(22kΩ + 2.2kΩ) = 81 V

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