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

• 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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