MCQ
The equivalent capacitance between $A$ and $B$ is


- A$\frac{C}{4}$
- B$\frac{{3C}}{4}$
- C$\frac{C}{3}$
- ✓$\frac{{4C}}{3}$


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$ x = 2 \sin \omega t \,;$ $ y = 2 \sin \left( {\omega t + \frac{\pi }{4}} \right)$
The path of the particle will be :


