A 1.5V battery is connected to a small light bulb with a resistance of 3.5Ω. What is the current in the bulb?

5 Answers

  • Ohm's Law: v = ir

    where

    v = voltage, Volts

    i = current, Amps

    r = resistance, Ohms

    i = v/r

    i = 1.5/3.5 = 0.43 Amps

  • From Ohm's regulation, V is without postpone proportional to I if different actual parts alongside with temperature are stored consistent. So, V=IR the place V is the p.d. around the bulb, I is the present which flows throughout the bulb and R is the resistance of the bulb. subsequently, I=V/R =4.7/7.9 I=595 mA.

  • If the bulb is ohmic (follows ohms law as maximum bulbs in the market do),

    V=IR

    I=V/R

    =1.5/3.5

    =3/7 ampere

  • Use ohms law. Voltage = Current*Resistance or V=IR

    solving for I ====> V/R=I

    I = 428mA or .428A

    hope this helps

  • Using Ohm’s law,

    V = IR

    I = V/R

    Where V = voltage, I = current, and R = resistance

    From the question,

    V = 1.5v

    R = 3.5Ω

    I = ?

    Substituting into the formula,

    I = V/R

    = 1.5/3.5

    = 0.429 ≈ 0.43Amp

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