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Question: A torch bulb is rated \( 5{\text{V}} \) and \( 500{\text{mA}} \) . Calculate (i) its power (ii) its ...

A torch bulb is rated 5V5{\text{V}} and 500mA500{\text{mA}} . Calculate (i) its power (ii) its resistance, and (iii) the energy consumed if this bulb is lighted for four hours.

Explanation

Solution

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=VIP = VI , here PP is the power dissipation across a resistance when a voltage VV is applied and a current of II flows through it.
V=IRV = IR , here VV is the voltage across a resistance RR and II is the current flowing through it.
E=PtE = Pt , here EE is the energy consumed due to a constant power PP in time tt .

Complete step by step answer
Let the resistance of the bulb be RR .
The rating of the bulb is 5V5{\text{V}} and 500mA500{\text{mA}} . This means that the bulb works when a voltage of 5V5{\text{V}} is applied across it and 500mA500{\text{mA}} of current flows through it.
(i) The power dissipated across a resistance is given by
P=VIP = VI
Substituting V=5VV = 5{\text{V}} , and I=500mAI = 500{\text{mA}} we get
P=5×500mWP = 5 \times 500{\text{mW}}
P=2500mW=2.5W\Rightarrow P = 2500{\text{mW}} = 2.5{\text{W}} …………………….(1)
Hence, the power dissipated across the torch bulb is equal to 2.5W2.5{\text{W}} .
(ii) From Ohm’s law we know that
V=IRV = IR
So the resistance is given by
R=VIR = \dfrac{V}{I}
Substituting V=5VV = 5{\text{V}} , and I=500mA=0.5AI = 500{\text{mA}} = 0.5{\text{A}} we get
R=50.5R = \dfrac{5}{{0.5}}
R=10Ω\Rightarrow R = 10\Omega
Hence, the resistance of the bulb is equal to 10Ω10\Omega .
(iii) We know that the energy is related to the power by
E=PtE = Pt …………………….(2)
According to the question, the time is t=4ht = 4h
We know that 1h=60×60s=3600s1h = 60 \times 60s = 3600s
So the time is
t=4×3600st = 4 \times 3600s
t=14400s\Rightarrow t = 14400s …………………….(3)
Substituting (1) and (3) in (2) we get
E=2.5×14400E = 2.5 \times 14400
E=36000J=36kJ\Rightarrow E = 36000{\text{J}} = 36{\text{kJ}}
Hence, the energy consumed by the bulb is equal to 36kJ{\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.