Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# A lamp is marked $60\,{\text{W}}$, $220\,{\text{V}}$. If it operates at $200\,{\text{V}}$, the rate of consumption energy will - - -.A. decreaseB. increaseC. remain unchangedD. first increase then decrease

Last updated date: 14th Jun 2024
Total views: 384.3k
Views today: 7.84k
Verified
384.3k+ views
Hint:Use the formula for the power. This formula gives the relation between the power, potential difference and the resistance. Using this formula, calculate the resistance of the circuit in the lamp. Use this value of the resistance to calculate the rate of energy consumption that is power for the new value of the potential difference. Compare the two values of the power to choose the correct option.

Formula used:
The power $P$ is given by
$P = \dfrac{{{V^2}}}{R}$ …… (1)
Here, $V$ is the potential difference and $R$ is the resistance.

The rate of the energy consumption is known as power.We have given that for potential difference of $220\,{\text{V}}$, the rate of energy consumption or power is $60\,{\text{W}}$.
${V_1} = 220\,{\text{V}}$
$\Rightarrow {P_1} = 60\,{\text{W}}$
We have asked to calculate the rate of energy consumption for the potential difference of $200\,{\text{V}}$.
${V_2} = 200\,{\text{V}}$

Let us first calculate the resistance ${R_1}$ for the first given pair of the potential difference and the rate of energy consumption.
Rewrite equation (1) for the power ${P_1}$.
${P_1} = \dfrac{{V_1^2}}{{{R_1}}}$
Rearrange the above equation for ${R_1}$.
${R_1} = \dfrac{{V_1^2}}{{{P_1}}}$
Substitute $220\,{\text{V}}$ for ${V_1}$ and $60\,{\text{W}}$ for ${P_1}$ in the above equation.
${R_1} = \dfrac{{{{\left( {220\,{\text{V}}} \right)}^2}}}{{60\,{\text{W}}}}$
$\Rightarrow {R_1} = 806.7\,\Omega$
Therefore, the resistance from the first pair of the power and potential difference is $806.7\,\Omega$.

We can suppose that the resistance of the circuit in the lamp remains the same for the changed value of the potential difference.Hence, the equation (1) for the changed value of the potential difference becomes
${R_1} = \dfrac{{V_2^2}}{{{P_2}}}$
Here, ${P_2}$ is the power of the lamp for the changed value of the potential difference.
Rearrange the above equation for ${P_2}$.
${P_2} = \dfrac{{V_2^2}}{{{R_1}}}$
Substitute $200\,{\text{V}}$ for ${V_2}$ and $806.7\,\Omega$ for ${R_1}$ in the above equation.
${P_2} = \dfrac{{{{\left( {200\,{\text{V}}} \right)}^2}}}{{806.7\,\Omega }}$
$\therefore {P_2} = 49.58\,{\text{W}}$
Hence, the new power is $49.58\,{\text{W}}$ which is less than the initial power.Therefore, the rate of the energy consumption will decrease.

Hence, the correct option is A.

Note: One can also solve the same question by another method. Once can use the same formula for the power which is the rate of consumption of energy in terms of potential difference and the resistance. Since we have taken the resistance constant for both the values of the potential difference, we can conclude that the power is directly proportional to the square of the potential difference. Hence, we can say that for the decreased value of the potential difference, the rate of energy consumption will also decrease.