Question

A $2W$ carbon resistor is colour-coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is
(A) $63mA$
(B) $0.4mA$
(C) $100mA$
(D) $20mA$

Answer 1

Hint
To find the maximum current passing through the resistor, we need to first calculate the value of the resistance using the colour codes. Then, we have to apply the formula $P = {I^2}R$ we can get the value of current.
Formula used: In this solution we will be using the following formula,
$P = {I^2}R$’
Where, $P$ is power, $I$ is the current flowing, and $R$ is the resistance.

Complete step by step answer
The table for the colour codes for carbon is:

LETTERS	COLOUR	FIGURE	MULTIPLIER
B	Black	0	${10^0}$
B	Brown	1	${10^1}$
R	Red	2	${10^2}$
O	Orange	3	${10^3}$
Y	Yellow	4	${10^4}$
G	Green	5	${10^5}$
B	Blue	6	${10^6}$
V	Violet	7	${10^7}$
G	Grey	8	${10^8}$
W	white	9	${10^9}$

Now, it is given in the question that the power of the carbon resistor is 2W. It is colour coded with green, black, red, and brown.
According to the table, green is 5, black is 0, red is ${10^2}$
So we get the resistance as $R = 50 \times {10^2}\Omega$
In the relation,
$P = {I^2}R$
From here we can rearrange this formula as,
${I^2} = \dfrac{P}{R}$
On taking root, $I = \sqrt {\dfrac{P}{R}}$
If $I$ is maximum, then the resistance should be minimum. Thus,
${R_{\min }} = 50 \times {10^2}\Omega$ ,
Thus, when we apply the values of $R$ and $P$ in the given formula,
We get,
${I_{\max }} = \sqrt {\dfrac{P}{R}}$
Substituting the values we get
$\Rightarrow {I_{\max }} = \sqrt {\dfrac{2}{{50 \times {{10}^2}}}}$
When we simplify, we get
$\Rightarrow {I_{\max }} = 2 \times {10^{ - 2}}A$
We can write this in milli-ampere as,
$\Rightarrow {I_{\max }} = 20mA$
Thus, we get the maximum current which can be passed through the resistor is $20mA$.
Hence, the correct answer is option (D).

Note
In the colour code for the resistor, we are given four colours. The fourth colour given as brown indicates the tolerance of the material. The colour brown means that the tolerance of that resistor is $\pm 1\%$.