A light bulb operates at a voltage V_0 and co | vedclass

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

Question

(D) The power consumed by a resistor (bulb) is given by the formula $P = \frac{V^2}{R}$,where $R$ is the resistance of the bulb.Initially,the bulb ope

Vedclass · Accepted Answer

$P = \left( \frac{V}{V_0} \right)^2 P_0$

Vedclass · Answer

(D) The power consumed by a resistor (bulb) is given by the formula $P = \frac{V^2}{R}$,where $R$ is the resistance of the bulb.Initially,the bulb operates at voltage $V_0$ and power $P_0$. Therefore,the resistance $R$ is given by:$R = \frac{V_0^2}{P_0} \quad \dots(1)$When the bulb is connected to a new voltage $V$,it consumes a new power $P$. Since the resistance $R$ of the bulb remains constant,we have:$P = \frac{V^2}{R}$Substituting the value of $R$ from equation $(1)$ into this expression:$P = \frac{V^2}{(V_0^2 / P_0)}$$P = \left( \frac{V^2}{V_0^2} \right) P_0$$P = \left( \frac{V}{V_0} \right)^2 P_0$

$A$ light bulb operates at a voltage $V_0$ and consumes power $P_0$. If the voltage is changed to $V$,what is the new power $P$ consumed by the bulb?

Similar Questions

An electric room radiator which operates at $225\, V$ has a resistance of $50\, \Omega$. The power of the radiator is approximately ............... $W$.

$A$ light bulb is rated at $200 \text{ W}$ for a $220 \text{ V}$ supply. Find the resistance of the bulb. (in $\Omega$)

How much electric energy is consumed by an average Indian household in one second?

$A$ heater draws a current of $2\,A$ when connected to a $250\,V$ source. The rate of energy dissipation is ............. $W$.

If the resistance of the filament increases with temperature,what will be the power dissipated in a $220\, V- 100\, W$ lamp when connected to a $110\, V$ power supply?

Vedclass Test Series

Exam Paper Generator

Online Exam Module