Problem — torch bulb

Problem

A bulb of resistance $R = 240\ \Omega$ is connected to a voltage $V = 12\ \text{V}$. Calculate the current, the power dissipated and the energy consumed in $10$ minutes of operation.

Bulb of resistance $R$ powered by battery $V$.

Solution

Current — from Ohm’s law: $i = \frac{V}{R} = \frac{12\ \text{V}}{240\ \Omega} = 0{,}050\ \text{A} = 50\ \text{mA}$

Power dissipated — the bulb converts all the electrical power into light and heat: $P = V\,i = 12\ \text{V} \cdot 0{,}050\ \text{A} = 0{,}60\ \text{W}$

Energy in $10$ minutes — with $t = 10\ \text{min} = 600\ \text{s}$: $E = P\,t = 0{,}60\ \text{W} \cdot 600\ \text{s} = 360\ \text{J}$

$\ev{i = 50\ \text{mA}, \quad P = 0{,}60\ \text{W}, \quad E = 360\ \text{J}}$

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Topics: Circuiti elettrici Concepts: Legge di Ohm · Potenza