Inductive Reactance

Using received values of 230 V at 50 Hz and a 100 Ω & 10 mH.

$$X_L=\mathrm{2\pi fL}$$ $$X_L=\mathrm{3.14\; \Omega }$$

Impedance

Using received values of 230 V at 50 Hz and a 100 Ω & 10 mH.

$$Z_{}=\sqrt{R^2+X^2}$$ $$Z_{}=\sqrt{{\mathrm{100}}^2+{\mathrm{3.14}}^2}$$ $$Z_{}=\sqrt{{\mathrm{10000}}^{}+{\mathrm{9.86}}^{}}$$ $$Z_{}=\sqrt{{\mathrm{10009.86}}^{}}$$ $$Z_{}=\mathrm{100.05\; \Omega }$$

Current

Using received values of 230 V at 50 Hz and a 100 Ω & 10 mH.

$$I_{}=\frac{V}{Z}$$ $$I_{}=\frac{\mathrm{230}}{\mathrm{100.05}}$$ $$I_{}=\mathrm{2.3\; A}$$

Voltages

Using received values of 230 V at 50 Hz and a 100 Ω & 10 mH.

Using Ohm's Law we can calculate the voltages present across each component as follows.

$$V_R=IxR$$ $$V_R=2.3x100=230V$$ $$V_L=IxL$$ $$V_L=2.3x3.14=7.222V$$

Unless one of either resistance or inductance is zero the sum of these two voltages does not add up to the supply voltage.

If we were to draw the two voltage values at right angles to each other then the resulting hypotenuse would equal the supply voltage (or very close to it allowing for rounding errors).

We'll now check.

$$V_{\mathrm{Supply}}=\sqrt{{V_{\mathrm{R}}}^2+{V_{\mathrm{L}}}^2}$$ $$V_{\mathrm{Supply}}=\sqrt{{\mathrm{230}}^2+{\mathrm{7.222}}^2}$$ $$V_{\mathrm{Supply}}=\sqrt{{\mathrm{52900}}^{}+{\mathrm{52.157284}}^{}}$$ $$V_{\mathrm{Supply}}=\sqrt{{\mathrm{52952.157284}}^{}}$$ $$V_{\mathrm{Supply}}=230.11V$$