In a circuit S(1) remains closed for a l...

In a circuit `S_(1)` remains closed for a long time and `S_(2)` remain open. Now `S_(2)` is closed and `S_(1)` is opened. Find out the `di//dt` in the right loop just after the moment.

Text Solution

Verified by Experts

Before `S_(2)` is closed and `S_(1)` is opened in the left part of the circuit `= varepsilon//R`. Now when `S_(2)` is closed and `S_(1)` is opened, current through the inductor cannot change suddenly, current `varepsilon//R` will continue to flow through the inductor.

Applying `KVL` in loop 1,
`L(di)/(dt) + (varepsilon)/(R ) (2R) + 4 varepsilon = 0 rArr (di)/(dt) = - (6 varepsilon)/(L)`

Topper's Solved these Questions

INDUCTANCE
CENGAGE PHYSICS|Exercise Exercises (subjective)|7 Videos
INDUCTANCE
CENGAGE PHYSICS|Exercise Exercises (single Correct )|65 Videos
INDUCTANCE
CENGAGE PHYSICS|Exercise Solved Examples|3 Videos
HEATING EFFECT OF CURRENT
CENGAGE PHYSICS|Exercise Thermal Power in Resistance Connected in Circuit|28 Videos
KINETIC THEORY
CENGAGE PHYSICS|Exercise Question Bank|31 Videos

The capacitor 'C' is initially unchanged. Switch S_(1) is closed for a long time while S_(2) remains open. Now at t = 0 S_(2) is closed while S_(1) is opened. All the batteries are ideal and connecting wires are resistanceless

At time t = 0, reading of ammeter is `(E)/(5R)`

At time t =0, reading of ammeter is zero

Heat developed till time `t = 5 RC l n 2` in resistance 3 R is `(9)/(40) CE^(2)`

After time `t gt 0` charge on the capacitor follows the equation `CEe^(-t//5RC)`

As shown in diagram initial charge on capacitor is zreo . Now switch S_(1) is closed & it remains closed for a long time but S_(2) remains open . Now switch S_(1) is opened & switch S_(2) is closed at t=0, then after closing the switch S_(2) :

Maximum current passing through inductor will be `(epsilon) sqrt((C)/(2L))`.

Maximum magnitude of rate of change of current through inductor will be `(epsilon)/(L)`.

Maximum charge on capacitor `C_(2)` will be `epsilonC`.

Maximum charge on capacitor `C_(2)` will be `(epsilonC)/(2)`.

In the circuit shows (Fig.) switches S_(1) and S_(3) have been closed for 1 s and S_(2) remained open. Just after 1 s , switch S_(2) is closed and S_(1) and S_(3) are opened. Find after that instant (t = 0) : the maximum current in the circuit containing inductor and capacitor (only S_(2) is closed)

`sqrt(3) (1 - (1)/(e))`

`sqrt(2) (1 - (1)/(e))`

`sqrt(3) (1 + (1)/(e))`

`sqrt(2) (1 + (1)/(e))`