Electro Magnetic Induction And Alternating Currents Question 413
Question: A 300 $ \Omega $ . resistor is connected in series with a parallel-plate capacitor across the terminals of a 50.0 Hz ac generator. When the gap between the plates is empty, its capacitance is $ 70/22\mu F $ . The ratio of the rms current in the circuit when the capacitor is empty to that when ruby mica of dielectric constant k = 5.0 is inserted between the plates, is equal to
Options:
0.1
0.3
0.6
2.9
Show Answer
Answer:
Correct Answer: B
Solution:
- $ i_{rms}=\frac{E_{0}}{\sqrt{2Z}}\Rightarrow \frac{{i_{rms,1}}}{{i_{rms,2}}}=\frac{Z_{2}}{Z_{1}}=\frac{\sqrt{R^{2}+{{( \frac{1}{\omega kC} )}^{2}}}}{\sqrt{R^{2}+{{( \frac{1}{\omega ,C} )}^{2}}}} $
Solving:
$ \Rightarrow \frac{{i_{rms,1}}}{{i_{rms,2}}}=\frac{1}{\sqrt{2}} $
BETA
