Two electric bulbs whose resistances are in the ratio of $1 : 2$ are connected in parallel to a constant voltage source. The powers dissipated in them have the ratio:

  • A
    $1:2$
  • B
    $1:1$
  • C
    $2:1$
  • D
    $1:4$

Explore More

Similar Questions

The potential difference $(V_A - V_B)$ between the points $A$ and $B$ in the given figure is ............ $V$.

In the given circuit,the current in each resistance is: (in $A$)

In the following circuit,the current through the resistor $R(=2 \Omega)$ is $I$ Amperes. The value of $I$ is

The smallest temperature difference that can be measured with a combination of a thermocouple of thermo $e.m.f.$ $30\,\mu V$ per degree and a galvanometer of $50\,\Omega$ resistance,capable of measuring a minimum current of $3 \times 10^{-7} \, A$ is ............... $^\circ C$.

Difficult
View Solution

$A$ cell of $emf$ $E$ and internal resistance $r$ is connected in series with an external resistance $nr$. Then,the ratio of the terminal potential difference to $emf$ is

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