In a series L-C-R circuit the voltage across resistance, capacitance and inductance is 10 V each. If the capacitance is short-circuited, the voltage across the inductance will be

Here E = X_L = X_C
($\because$ voltage across them is same)
Total voltage in the circuit
$=\mathrm{I}{\left[{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2\right]}^{1/2}=\mathrm{IR}=10\mathrm{V}$
When capacitor is short-circuited
$\mathrm{I}=\frac{10}{{\left({\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2\right)}^{1/2}}=\frac{10}{\sqrt{2}\cdot \mathrm{R}}$
$\therefore$ Potential drop across inductance
$={\mathrm{IX}}_{\mathrm{L}}=\mathrm{IR}=(10/\sqrt{2})\mathrm{V}$