Home
Class 12
PHYSICS
In the circuit shown in Figure, the sour...

In the circuit shown in Figure, the source has a rating of 15 V, 100 Hz. The resistance R is 3 `Omega` and the reactance of the capacitor is 4 `Omega`. It is known that the box certainly contains one or more element (resistance, capacitance or inductance). Which element/s are present inside the box?

Text Solution

Verified by Experts

The correct Answer is:
The box has L and C in series
Promotional Banner

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

In the circuit shown, the cell is ideal , with emf= 15 V . Ecah resistance is of 3 Omega. the potential difference across the capacitor is

In the circuit shown the cell is idea with end =15V each resistance is of 3Omega The potential difference the capacitor is

A series combination of R,LC is connected to an a.c source. If the resistance is 3 Omega and the reactance is 4 Omega , the power factor of the circuit is

In a circuit shown in figure the battery emf is equal to E=2.0 V , its internal resistance is r=9.0 Omega , the capacitance of the capacitor is C=10 mu F , the coil inductance is L=100 m H , and the resistance is R=100 Omega . At a certain momenta the switch Sw was disconneted. Find the energy of oscillations in the circuit (b) t=0.30s after the switch was disconnected,

In the circuit shown in figure 3.242 capacitor is charged with 50muC charge. Find the current in 4Omega resistance just after switch S is closed.

In a seris LCR circuit, the inductive reactance (X_(L)) is 10 Omega and the capacitive reactance (X_(C)) is 4 Omega . The resistance (R) in the circuit is 6 Omega. The power factor of the circuit is :

The circuit shown in the figure contains two diodes each with a forward resistance of 50 Omega and with infinite backward resistance. If the battery is 6 V, the current through the 100 Omega resistance (in ampere) is

In the circuit shown in Fig. the emf of the sources is equal to xi = 5.0 V and the resistances are equal to R_(1) = 4.0 Omega and R_(2) = 6.0 Omega . The internal resistance of the source equals R = 1.10 Omega . Find the currents flowing through the resistances R_(1) and R_(2) .