MCQ
In the given circuit what is the potential drop across resistance?
- A40 V
- B80 V
- C120 V
- Dzero
✓
Answer
- 120 V
Explanation:
At resonance the voltage across L and C are same but opposite.
So, at resonance $\mid\text{V}_\text{L}\mid=\mid\text{V}_\text{C}\mid$
$\therefore\text{V}_\text{R}=\text{V}_\text{app}=120\text{V}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Field lines start from____ charges and ends at _____ charges.
Consider a sample of a pure beta-active material:
- All the beta particles emitted have the same energy.
- The beta particles originally exist inside the nucleus and are ejected at the time of beta decay.
- The antineutrino emitted in a beta decay has zero mass and hence zero momentum.
- The active nucleus changes to one of its isobars after the beta decay.
A transverse wave travels along the Z-axis. The particles of the medium must move:
- Along the Z-axis
- Along the X-axis
- Along the Y-axis
- In the X-Y plane.
Binding energy of a nucleus is of the order of.
Resistance as shown in figure is negative at
|
(a) A |
(b) B |
(c) C |
(d) None of these |
When two bar magnets are coaxially placed at some distance from each other, then no torque works because:
The conductivity of a metal decreases with the increase in temperature on account of:
If we express the energy of a photon in KeV and the wavelength in angstroms, then energy of a photon can be calculated from the relation
|
(a) E = 12.4 hv |
(b) E = 12.4 h/ |
(c) E = 12.4/ |
(d) E = hv |
The momentum of a photon of wavelength $\lambda$ is:
→$\lambda$ ray of light passes through a prism of refractive index $\sqrt{2}$ in such a way that its angle of incidence is twice of the angle of refraction and there is minimum deviation in the ray. The angle of the prism will be :
→