The resistance of the filament of an electric | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The rated power is $P_1 = 100\,W$ at $V_1 = 220\,V$. The resistance at this temperature is $R_1 = \frac{V_1^2}{P_1}$.When connected to $V_2 = 220

Vedclass · Accepted Answer

Between $100 	imes (0.8)^2\,W$ and $100 	imes 0.8\,W$

Vedclass · Answer

(D) The rated power is $P_1 = 100\,W$ at $V_1 = 220\,V$. The resistance at this temperature is $R_1 = \frac{V_1^2}{P_1}$.When connected to $V_2 = 220 	imes 0.8\,V$,the new power is $P_2 = \frac{V_2^2}{R_2}$.Taking the ratio: $\frac{P_2}{P_1} = \frac{V_2^2}{V_1^2} 	imes \frac{R_1}{R_2} = (0.8)^2 	imes \frac{R_1}{R_2}$.Since the voltage decreases,the temperature of the filament decreases,which causes the resistance to decrease $(R_2  1$.This implies $P_2 > (0.8)^2 	imes P_1 = 100 	imes (0.8)^2\,W$.Also,since $P = VI$,and both $V$ and $I$ decrease compared to the rated values,the power $P_2$ must be less than the power calculated if resistance remained constant at $R_1$ (which would be $100 	imes 0.8\,W$).Thus,the actual power $P_2$ lies between $100 	imes (0.8)^2\,W$ and $100 	imes 0.8\,W$.

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $220\,V$ and $100\,W$ is connected to a $(220 \times 0.8)\,V$ source,then the actual power consumed would be:

Similar Questions

Consider the four circuits shown in the figure given below. In which circuit is the power dissipated the greatest? (Neglect the internal resistance of the power supply)

The effective resistance between points $P$ and $Q$ of the electrical circuit shown in the figure is

The potential difference across the terminals of the battery shown in the figure is .................... $V$. ($E = 10 \ V$,$r = 1 \ \Omega$,external resistance $R = 4 \ \Omega$) (in $V$)

The current flowing through the $1 \Omega$ resistor is $\frac{n}{10} \ A$. The value of $n$ is. . . . . . .

In the given circuit,the value of the current $I$ is: (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module