Question

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

Answer 1

Hint : To solve this question, we need to use the formula for the power dissipation across a resistor. The value of resistance can be found out by using the Ohm’s law. And the energy can be found out from the value of the power.

Formula used: The formulae used to solve this question are given by
 $ P = VI $ , here $ P $ is the power dissipation across a resistance when a voltage $ V $ is applied and a current of $ I $ flows through it.
 $ V = IR $ , here $ V $ is the voltage across a resistance $ R $ and $ I $ is the current flowing through it.
 $ E = Pt $ , here $ E $ is the energy consumed due to a constant power $ P $ in time $ t $ .

Complete step by step answer
Let the resistance of the bulb be $ R $ .
The rating of the bulb is $ 5{\text{V}} $ and $ 500{\text{mA}} $ . This means that the bulb works when a voltage of $ 5{\text{V}} $ is applied across it and $ 500{\text{mA}} $ of current flows through it.
(i) The power dissipated across a resistance is given by
 $ P = VI $
Substituting $ V = 5{\text{V}} $ , and $ I = 500{\text{mA}} $ we get
 $ P = 5 \times 500{\text{mW}} $
 $ \Rightarrow P = 2500{\text{mW}} = 2.5{\text{W}} $ …………………….(1)
Hence, the power dissipated across the torch bulb is equal to $ 2.5{\text{W}} $ .
(ii) From Ohm’s law we know that
 $ V = IR $
So the resistance is given by
 $ R = \dfrac{V}{I} $
Substituting $ V = 5{\text{V}} $ , and $ I = 500{\text{mA}} = 0.5{\text{A}} $ we get
 $ R = \dfrac{5}{{0.5}} $
 $ \Rightarrow R = 10\Omega $
Hence, the resistance of the bulb is equal to $ 10\Omega $ .
(iii) We know that the energy is related to the power by
 $ E = Pt $ …………………….(2)
According to the question, the time is $ t = 4h $
We know that $ 1h = 60 \times 60s = 3600s $
So the time is
 $ t = 4 \times 3600s $
 $ \Rightarrow t = 14400s $ …………………….(3)
Substituting (1) and (3) in (2) we get
 $ E = 2.5 \times 14400 $
 $ \Rightarrow E = 36000{\text{J}} = 36{\text{kJ}} $
Hence, the energy consumed by the bulb is equal to $ {\text{36kJ}} $ .

Note
We should not forget to convert the values of the quantities given into their respective SI units. For example, the current is given in milliamperes, also the time is given in hours. So these are supposed to be converted into SI units.