Electronical engineering question?

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  • 3 months ago

    V=4+12=16v

    I2=4/500=8mA

    R3=12/.008=1500Ω

    I1=18.7-8=10.7mA

    R1=16/10.7mA=1,397.5Ω

  • 3 months ago

    The battery voltage is 4+12 = 16 volts

    I2 = 4/0.5 = 8 mA

    R3 = 12/8 = 1.5 kΩ

    I1 = 18.7–8 = 10.7 mA

    R1 = 16/10.7 = 1.50 kΩ

    Ohm's law: voltage in volts = current in mA x resistance in kΩ

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