The circuit is closed for long time so that the capacitors are fully charged.Switch is opened at a certain instant.The heat developed in resistor `R_(2)` after that instant is `K times10^(-4) J.` Then find the value

In the circuit shown in fig. when the switch is closed, the capacitor charges with a time constant

At any time S1 switch is opened and S2 is closed then find out heat generated in circuit.

In the circuit shown, the capacitor is initially uncharged. The switch S is closed at t = 0. Time after which voltage across capacitor and resistor are equal is:

In the circuit shown, the capacitor C_1 is initially charged with charge q_0 . The switch S is closed at time t = 0 . The charge on C_2 after time t is

In the circuit in Fig. switch S_(1) was closed for a long time . At time t = 0 the switch is opened. Find the maximum potentail difference across the plates of the capacitor after the switch is opened.

In the LR circuit the switch S was closed for a long time. The ideal cell in the circuit has emf E volts.Later on the switch is opened. The current in the resistor of resistance R ohms just after the switch was opened is (XE)/(4R) , then x is

At any time S_(1) switch is opened and S_(2) is closed then find out heat generated in circuit.