Question

If $R_1$and $R_2$ be the resistances of the filaments of $200W$ and $100W$ electric bulbs operating on $220V$, then

• A. $R_1$ is equal to $R_2$
• B. $R_1$ is twice that of $R_2$
• C. $R_2$ is twice that of $R_1$
• D. None of these

Answer 1

The correct option is C

$R_2$ is twice that of $R_1$

Step 1: Analyzing the given data:

Given:

Power of bulb $1$ $\left(P_1\right)$ $=200W$

Power of bulb $2$ $\left(P_2\right)$ $=100W$

Potential Difference $\left(V\right)=220V$

Find:

Resistances ($R_1$ and $R_2$)

Step 2: Relation between power and resistance:

Electric power is given by:

$P=VI$

According to Ohm’s Law:

$V=IR$

$\therefore I=\frac{V}{R}$

Substituting the above equation in $P$:

$\therefore P=\frac{V^2}{R}$

Step 3: Determining resistance $R_1$ and $R_2$

For $R_1$:

$P_1=\frac{V^2}{R_1}$

$\therefore R_1=\frac{V^2}{P_1}$

For $R_2$:

$P_2=\frac{V^2}{R_2}$

$\therefore R_2=\frac{V^2}{P_2}$

Step 4: Calculating the ratio of resistances:

Consider, $\frac{R_1}{R_2}=\frac{\frac{V^2}{P_1}}{\frac{V^2}{P_2}}$

Simplifying the above equation, we get:

$\frac{R_1}{R_2}=\frac{P_2}{P_1}$

Substituting the given values above:

$\therefore \frac{R_1}{R_2}=\frac{100}{200}$

$\therefore \frac{R_1}{R_2}=\frac{1}{2}$

Thus, $R_2=2R_1$

$\Rightarrow$$R_2$is twice that of $R_1$

Hence, option (C) is the correct answer.