Solve

Guides

0

You visited us 0 times! Enjoying our articles? Unlock Full Access!

Question

21. A capacitor of capacitance C = 3 uF is first charged by connecting it across a 10 V battery by closing key Kı. It is then allowed to get discharged through 292 and 422 resistors by closing the key K2. The total energy dissipated in the 212 resistor is equal to TO (A) 0.5 m) (C) 0.15 m) + (B) 0.05 m) (D) none of these 42

Open in App

Solution

Verified by Toppr

Was this answer helpful?

0

Similar Questions

Q1

Three capacitors, each of capacitance C are connected to a battery of voltage V. When key K is closed, the charge which will flow through the battery is -

View Solution

Q2

A charged capacitor is allowed to discharge through a resistor by closing the key at the instant

t = 0

. At J the instant

t = (l n 4) μ s

, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal to

View Solution

Q3

By evaluating ∫ i²Rdt, show that when a capacitor is charged by connecting it to a battery through a resistor, the energy dissipated as heat equals the energy stored in the capacitor.

View Solution

Q4

Three capacitors, each of capacitance

4 μ F

are connected to a battery of voltage

10 V

. When key

K

is closed, the charge which will flow through the battery is

View Solution

Q5

Capacitor

C_{1}

of capacitance

1 μ F

and capacitor

C_{2}

of capacitance

2 μ F

are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time

t = 0

.

View Solution