Explore More

Similar Questions

Four capacitors with capacitances $C_1 = 1\,\mu F, C_2 = 1.5\,\mu F, C_3 = 2.5\,\mu F$ and $C_4 = 0.5\,\mu F$ are connected as shown and are connected to a $30\,V$ source. The potential difference between points $a$ and $b$ is.....$V$.

In $Millikan's$ oil drop experiment,a charge $Q$ is held stationary between two plates under a potential difference of $2400\, V$. If a second drop with half the radius is to be held stationary using a potential difference of $600\, V$,then the charge on the second drop is:

$A$ parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $E_2 / E_1$ is

Three capacitors $A, B$ and $C$ are connected with a battery of $emf$ $\varepsilon$. All capacitors are identical initially. If a dielectric slab is inserted between the plates of capacitor $A$ slowly with the help of an external force,then:

Find the equivalent capacitance between $A$ and $B$.

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