Given:
\[C = 2 \, \mu\text{F}, \quad E = 110\sqrt{2} \sin(100t).\]
The capacitive reactance is:
\[X_C = \frac{1}{\omega C},\]
where $\omega = 100 \, \text{rad/s}$ and $C = 2 \times 10^{-6} \, \text{F}$.
Substitute:
\[X_C = \frac{1}{100 \cdot 2 \times 10^{-6}} = \frac{1}{2 \times 10^{-4}} = 5000 \, \Omega.\]
The peak current is:
\[i_0 = \frac{E_0}{X_C},\]
where $E_0 = 110\sqrt{2} \, \text{V}$.
Substitute:
\[i_0 = \frac{110\sqrt{2}}{5000}.\]
The RMS value of current is:\[i_{\text{rms}} = \frac{i_0}{\sqrt{2}}.\]
Substitute:\[i_{\text{rms}} = \frac{110\sqrt{2}}{5000\sqrt{2}} = \frac{110}{5000}.\]
Simplify:\[i_{\text{rms}} = \frac{110}{5000} \, \text{A} = 22 \, \text{mA}.\]
Thus, the RMS value of current in the circuit is:\[i_{\text{rms}} = 22 \, \text{mA}.\]
Draw the plots showing the variation of magnetic flux φ linked with the loop with time t and variation of induced emf E with time t. Mark the relevant values of E, φ and t on the graphs.
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to: