Question

A torch bulb is rated \[2.5\,{\text{V}}\] and \[750\,{\text{mA}}\]. Calculate:
1) Its power (in \[{\text{W}}\] )
2) Its resistance (in \[\Omega \] ) and
3) The energy consumed (in \[{\text{J}}\] ) if this bulb is lighted for four hours.

Answer 1

Hint: First of all, we will use the formula which gives power equals to the product of current and potential difference. Then we will use Ohm's law to find the resistance. We will convert the given time into seconds and find the energy which is given by the product of power and time in seconds.

Complete step by step answer:
In the given question, we are supplied with the following data:
The rated voltage of the bulb is \[2.5\,{\text{V}}\] .
The rated current of the bulb is \[750\,{\text{mA}}\] .
We are asked to find the power of the bulb in watts, its resistance in ohms and the energy consumed by the bulb in joules.

1)Since, we are given the rated current of the bulb is \[750\,{\text{mA}}\] and the rated voltage of the bulb is \[2.5\,{\text{V}}\] .
We know,
\[1\,{\text{mA}} = 1 \times {10^{ - 3}}\,{\text{A}}\]
So,
$750\,{\text{mA}} = 750 \times {10^{ - 3}}\,{\text{A}} \\
750\,{\text{mA}} = 0.750\,{\text{A}} \\$
To calculate the power of the bulb, we will use the formula, as given below:
\[P = V \times i\] …… (1)
Where,
\[P\] indicates power of the bulb.
\[V\] indicates rated voltage of the bulb.
\[i\] indicates rated current of the bulb.
Now by substituting the required values in the equation (1), we get:
$P = V \times i \\
P = 2.5 \times 0.750\,{\text{W}} \\
P = {\text{1}}{\text{.875}}\,{\text{W}} \\$
Hence, the power of the bulb is found to be \[{\text{1}}{\text{.875}}\,{\text{W}}\] .

2) To find the resistance, we will directly apply the Ohm’s law, which is given by:
\[V = iR\] …… (2)
Where,
\[V\] indicates potential difference.
\[i\] indicates rated current of the bulb.
\[R\] indicates resistance of the bulb.
Substituting the required values in the equation (2), we get:
$V = iR \\
R = \dfrac{V}{i} \\
R = \dfrac{{2.5}}{{0.75}}\,\Omega \\
R = 3.33\,\Omega \\$
Hence, the resistance of the bulb is \[3.33\,\Omega \] .

3) To find the energy consumed by the bulb in joules, we will first convert the given duration of time into seconds.
$ 4\,{\text{h}} \\
= 4 \times 60\,\min \\
= 4 \times 60 \times 60\,{\text{s}} \\
= 14400\,{\text{s}} \\$
Energy is given by the formula as given below:
\[E = P \times t\] …… (3)
Where,
\[E\] indicates energy consumed.
\[P\] indicates power.
\[t\] indicates time for which the bulb was lighted.
Substituting the required values in the equation (3), we get:
$E = P \times t \\
E = 1.875 \times 14400\,{\text{J}} \\
E = 27000\,{\text{J}} \\$
Hence, the energy consumed by the bulb is \[27000\,{\text{J}}\] .

Note:
While solving this problem, always use the S.I units. The energy can also be conveyed in kilowatt hour, which is for commercial use. In that case, we need to convert the power from watt to kilowatts and the time will be in hours format.