A heater is marked 1.8 kW, 230 V. The mains has a frequency of 50 Hz and the 230 V is an rms value…

Step 1. The resistance. We want $R$, so $R = \dfrac{V_{rms}^2}{P} = \dfrac{(230\ \text{V})^2}{1800\ \text{W}} = 29.4\ \Omega$.

Step 2. The peak current. $I_{rms} = \dfrac{P}{V_{rms}} = \dfrac{1800\ \text{W}}{230\ \text{V}} = 7.83$ A. So $I_m = \sqrt{2}\, I_{rms} = 1.414 \times 7.83\ \text{A} = 11.1$ A.

Step 3. The greatest power. The peak voltage is $V_m = 1.414 \times 230\ \text{V} = 325$ V. At the peak, $p = V_m I_m = 325\ \text{V} \times 11.1\ \text{A} = 3.6$ kW. This is twice the rated power of $1.8$ kW.