1. **State the problem:** We need to calculate the total impedance $Z$ of a series circuit containing a resistor $R = 33\ \Omega$ and a capacitor with capacitance $C = 200 \times 10^{-6}\ F$ (or $200\ \mu F$) supplied by an AC voltage. 2. **Relevant formula:** The total impedance in a series RC circuit is given by $$Z = \sqrt{R^2 + X_c^2}$$ where $X_c$ is the capacitive reactance. 3. **Calculate capacitive reactance $X_c$:** The capacitive reactance is $$X_c = \frac{1}{2 \pi f C}$$ where $f$ is the frequency of the AC supply. 4. **Find the frequency $f$ from the graph:** The waveform completes 2 cycles in 0.04 seconds, so the period $T$ is $$T = \frac{0.04}{2} = 0.02\ s$$ Frequency is $$f = \frac{1}{T} = \frac{1}{0.02} = 50\ Hz$$ 5. **Calculate $X_c$:** $$X_c = \frac{1}{2 \pi \times 50 \times 200 \times 10^{-6}} = \frac{1}{2 \pi \times 50 \times 0.0002}$$ $$= \frac{1}{0.062832} \approx 15.92\ \Omega$$ 6. **Calculate total impedance $Z$:** $$Z = \sqrt{33^2 + 15.92^2} = \sqrt{1089 + 253.5} = \sqrt{1342.5} \approx 36.64\ \Omega$$ **Final answer:** The total impedance in the circuit is approximately $36.64\ \Omega$.