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Kelsie Sawayn

Feb 20, 2025

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

M
Madie Hilpert

Feb 20, 2025

Ohm's Law: v = irwherev = voltage, Voltsi = current, Amps r = resistance, Ohmsi = v/ri = 1.5/3.5 = 0.43 Amps... Show More
M
Madie Hilpert

Feb 20, 2025

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.... Show More
M
Madie Hilpert

Feb 20, 2025

If the bulb is ohmic (follows ohms law as maximum bulbs in the market do), V=IRI=V/R=1.5/3.5=3/7 ampere... Show More
M
Madie Hilpert

Feb 20, 2025

Use ohms law. Voltage = Current*Resistance or V=IRsolving for I ====> V/R=II = 428mA or .428Ahope this helps... Show More
M
Madie Hilpert

Feb 20, 2025

Using Ohm’s law,V = IRI = V/RWhere V = voltage, I = current, and R = resistanceFrom the question,V = 1.5vR = 3.5ΩI = ?Substituting into the formula,I = V/R = 1.5/3.5 = 0.429 ≈ 0.43Amp... Show More

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