A direct current of 4 A and an alternating current of 4 A flows through two identical resistors. The ratio of heat produced in the two resistances in same time interval will be

At resonance V_L = V_C

_{$\Rightarrow cos?=\frac{V_R}{\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}}=1$}

$X_C=\frac{1}{\omega C}$

On increasing $\omega$ $$ XC decreases so Z decreases, current in circuit increases.

$E=\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}$

$E=\sqrt[]{{\left(60\right)}^2+{\left(100-20\right)}^2}$

E = 100V

R = $2\Omega$

L = 2 mH

E = 9V

$i=\frac{\epsilon }{2R}=\frac{9v}{4\Omega }=2.25A$

Just after the switch ‘S’ is closed, the inductor acts as open circuit.

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5\times 10^{-3}\right)\times \left(200\times 10^{-6}\right)}}$

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5\times 10^{2}Hz$ $\left[Taking\pi =\sqrt[]{10}\right]$