Physics - Current Electricity Question with Solution | TestHub

Let $I_1$ and $I_2$ be currents through $B_1$ and $B_2$ then $I_1\times \sqrt{R^2\times X_L^2}=220$ $\frac{I_2}{I_1}=\frac{\sqrt{R^2\times X_C^2}}{\sqrt{R^2+X_L^2}}=\frac{\sqrt{R^2+{\left(\frac{1}{c\omega }\right)}^2}}{\sqrt{R^2+L^2{\omega }^2}}$ $=\frac{\sqrt{R^2+{\left(\frac{1}{500\times 10^{-6}\times 2\pi \times 50}\right)}^2}}{\sqrt{R^2+{\left(10\times 10^{-3}\times 2\pi \times 50\right)}^2}}$ $=\sqrt{R^2+40}/\sqrt{R^2+9.87},I_2>I_1$ Bulb $B_2$ will be brighter. As frequency and increases, $X_C$ decreases $X_L$ increases. $I_2$ becomes less and $I_1$ increases.

∴ brightness of $B_1$ will increase and that of $B_2$ decreases.