In a series $LCR$ circuit, the voltage across resistance, capacitance and inductance is $10\mathrm{ }\mathrm{V}$ each. If the capacitor is removed, the voltage across the inductance will be

Given, $V_{\mathrm{R}}=V_{\mathrm{L}}=V_{\mathrm{C}}=10 \mathrm{V}$

This means the LCR circuit is in resonance and the reactance of inductor and capacitor is equal to the resistance of the resistor.

$\Rightarrow V_{\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\mathrm{l}\mathrm{y}}=V_{\mathrm{R}}=10\mathrm{ }\mathrm{V}$

After removing capacitor

$V_{\mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\mathrm{l}\mathrm{y}}=\sqrt{V_{\mathrm{R}}^2+V_{\mathrm{L}}^2}$

Since, $V_{\mathrm{R}}=V_{\mathrm{L}}$

$\Rightarrow 10=\sqrt{{V_{\mathrm{L}}}^2+{V_{\mathrm{L}}}^2}$

$\Rightarrow 2{V_{\mathrm{L}}}^2=100$

$V_{\mathrm{L}}=\frac{10}{\sqrt{2}}\mathrm{ }\mathrm{V}$