Question

A torch bulb is rated $5V$ and $500mA$ . Calculate $\left( i \right)$ its power $\left( {ii} \right)$its resistance and $\left( {iii} \right)$the energy consumed if the bulb is lighted for $4$ hours.

Answer 1

Hint This problem can be solved by using the electric current and its application. Resistance can be obtained by using ohm's law which states that electric current passing through any conductor is directly proportional to the potential difference across its two ends.
Formula used:
$V = IR$(Ohm's law)
$P = VI$ (Electrical power)
$E = Pt$ (Electrical energy)

Complete Step by step solution
Here given that bulb has potential difference $V = 5V$and current $I = 500mA$. We will separately find all the quantities by using the given quantities. We will proceed with
$\left( i \right)$Power $(P)$
The electrical power $(P)$ is a product potential difference $(V)$ and electric current$(I)$. Electrical power is the rate of electrical energy transferred to electric current per unit time. Its SI unit is watt or joules per second. Hence
$P = VI$
$\Rightarrow P = 5V \times 0.5A$
$\Rightarrow P = 2.5W$ (In watts)
$\left( {ii} \right)$Resistance $(R)$
Ohm's law states that electric current passing through any conductor is directly proportional to the potential difference across its two ends.
$V \propto I$
$\Rightarrow V = RI$
where $R$ is constant of proportionality and it is known as resistance.
According to Ohm's law, resistance $(R)$ is a ratio of potential difference $(V)$ to electric current$(I)$.
$V = IR$
$\Rightarrow R = \dfrac{V}{I}$
Substituting the values of potential difference $(V)$ and electric current $(I)$we get
$R = \dfrac{{5V}}{{0.5A}}$
$\therefore R = 10\Omega$
$\left( {iii} \right)$Energy consumed by the bulb in $t = 4hrs = 14400s$
Electrical energy is a product of electrical power$(P)$ and time$(t)$. Electrical energy is the energy obtained by electrical potential energy. Its SI unit is Joules.
$E = Pt$
$\Rightarrow E = 2.5W \times 14400s = 36000J$
$\therefore E = 36kJ$
Hence for the given torch bulb rated $5V$and $500mA$ its $(i)$ Power$P = 2.5W$, $(ii)$Resistance $R = 10\Omega$
$(iii)$Electrical energy consumed when the bulb is lighted for $4hrs$ is $E = 36kJ$.

Note While solving the numerical ensure each physical quantity is in its SI units. If not then proceed with converting it first. One should not get confused about electrical energy with electrical power. Electrical power indicates the rate at which energy is consumed in a circuit while electric energy is derived by the electrical potential energy.