Two electric bulbs rated P_1 text{ watt}, V t | vedclass

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

Question

(C) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.For the first bulb

Vedclass · Accepted Answer

$(P_1 + P_2) 	ext{ watt}$

Vedclass · Answer

(C) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance of the bulb.For the first bulb,$R_1 = \frac{V^2}{P_1}$.For the second bulb,$R_2 = \frac{V^2}{P_2}$.When connected in parallel,the total power $P_{	ext{total}}$ is the sum of the individual powers consumed by each bulb.Since the voltage $V$ applied is the same as the rated voltage,each bulb consumes its rated power.Therefore,$P_{	ext{total}} = P_1 + P_2$.

Two electric bulbs rated $P_1 \text{ watt}, V \text{ volt}$ and $P_2 \text{ watt}, V \text{ volt}$ are connected in parallel and $V \text{ volt}$ is applied to them. The total power consumed will be

Similar Questions

The biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is

Which of the following is the strongest Bronsted base?

An electron moves in a circular orbit with a uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is proportional to

The reaction $RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO + HCl$ is known as:

$A$ sphere $P$ of mass $m$ moving with velocity $v$ undergoes an oblique and perfectly elastic collision with an identical sphere $Q$ initially at rest. The angle $\theta$ between the velocities of the spheres after the collision shall be ............... $^o$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module