(A) The resistance of the bulbs are given by $R_1 = \frac{V^2}{P_1}$ and $R_2 = \frac{V^2}{P_2}$.When connected in parallel,the total power $P_p$ cons

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

(A) The resistance of the bulbs are given by $R_1 = \frac{V^2}{P_1}$ and $R_2 = \frac{V^2}{P_2}$.When connected in parallel,the total power $P_p$ consumed by the circuit is given by $P_p = \frac{V^2}{R_{eq}}$.For parallel connection,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.Substituting the values,we get $\frac{P_p}{V^2} = \frac{P_1}{V^2} + \frac{P_2}{V^2}$.Therefore,$P_p = P_1 + P_2$.

Class 12 Physics Current Electricity Electrical Energy and Power

Easy

Two electric bulbs have ratings $P_1 \text{ watt}, V \text{ volt}$ and $P_2 \text{ watt}, V \text{ volt}$. If they are connected in parallel to a $V \text{ volt}$ supply,the total power consumed will be:

A
$P_1 + P_2 \text{ watt}$
B
$\sqrt{P_1 P_2} \text{ watt}$
C
$\frac{P_1 P_2}{P_1 + P_2} \text{ watt}$
D
$\frac{P_1 + P_2}{P_1 P_2} \text{ watt}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Current Electricity

Subtopic

Electrical Energy and Power

The resistance of a device component decreases as the current through it increases and it is described by the relation,$R = \frac{0.2 I}{I-4}$,where $I$ is the current. Determine the minimum power delivered. (Assume $I > 4$) (in $W$)

DifficultTS EAMCET 2019

View Solution

An electric bulb rated $50 \, W-200 \, V$ is connected across a $100 \, V$ supply. The power dissipation of the bulb is: (in $ \, W$)

DifficultJEE MAIN 2024

View Solution

When two electric bulbs of $40\, W$ and $60\, W$ are connected in parallel,then:

Easy

View Solution

How much work is required to carry a $6 \mu C$ charge from the negative terminal to the positive terminal of a $9 V$ battery?

Easy

View Solution

An electric kettle takes $4 \ A$ current at $220 \ V$. If the entire electric energy is converted into heat energy, then the time (in minutes) taken to increase the temperature of $1 \ kg$ of water from $34^{\circ} C$ to $100^{\circ} C$ is

MediumAP EAMCET 2025

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Two electric bulbs have ratings $P_1 \text{ watt}, V \text{ volt}$ and $P_2 \text{ watt}, V \text{ volt}$. If they are connected in parallel to a $V \text{ volt}$ supply,the total power consumed will be:

Similar Questions

The resistance of a device component decreases as the current through it increases and it is described by the relation,$R = \frac{0.2 I}{I-4}$,where $I$ is the current. Determine the minimum power delivered. (Assume $I > 4$) (in $W$)

An electric bulb rated $50 \, W-200 \, V$ is connected across a $100 \, V$ supply. The power dissipation of the bulb is: (in $ \, W$)

When two electric bulbs of $40\, W$ and $60\, W$ are connected in parallel,then:

How much work is required to carry a $6 \mu C$ charge from the negative terminal to the positive terminal of a $9 V$ battery?

An electric kettle takes $4 \ A$ current at $220 \ V$. If the entire electric energy is converted into heat energy, then the time (in minutes) taken to increase the temperature of $1 \ kg$ of water from $34^{\circ} C$ to $100^{\circ} C$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module