A 750 Hz, 20 V(rms) source is connected to a resistance of 100 $$\Omega$$, an inductance of 0.1803 H and a capacitance of 10 $$\mu$$F all in series. The time in which the resistance (heat capacity 2 J/°C) will get heated by 10°C (assume no loss of heat to the surroundings) is close to:

The source has frequency $$f = 750 \text{ Hz}$$ and rms voltage $$V_{\text{rms}} = 20 \text{ V}$$.

First we convert frequency to angular frequency using the relation $$\omega = 2\pi f$$. Hence

$$\omega = 2\pi \times 750 \; \text{rad s}^{-1} = 1500\pi \; \text{rad s}^{-1} \approx 4712 \; \text{rad s}^{-1}.$$

For the series combination we need the reactances. The inductive reactance is defined by the formula $$X_L = \omega L$$. Substituting $$L = 0.1803\ \text{H}$$,

$$X_L = 4712 \times 0.1803 \; \Omega \approx 849.6 \; \Omega.$$

The capacitive reactance is given by $$X_C = \dfrac{1}{\omega C}$$. With $$C = 10\ \mu\text{F} = 10 \times 10^{-6}\ \text{F},$$

$$X_C = \dfrac{1}{4712 \times 10 \times 10^{-6}} \; \Omega = \dfrac{1}{0.0471239} \; \Omega \approx 21.23 \; \Omega.$$

The net reactance of the series circuit is therefore

$$X = X_L - X_C = 849.6 \; \Omega - 21.23 \; \Omega \approx 828.4 \; \Omega.$$

The magnitude of the total impedance of a series $$RLC$$ circuit is obtained from the relation

$$Z = \sqrt{R^2 + X^2}.$$

Putting $$R = 100\ \Omega$$ and $$X = 828.4\ \Omega$$,

$$Z = \sqrt{(100)^2 + (828.4)^2} = \sqrt{10000 + 686263} = \sqrt{696263} \approx 834.5\ \Omega.$$

The rms current in the circuit is given by Ohm’s law $$I_{\text{rms}} = \dfrac{V_{\text{rms}}}{Z}$$. Using $$V_{\text{rms}} = 20\ \text{V},$$

$$I_{\text{rms}} = \dfrac{20}{834.5}\ \text{A} \approx 2.396 \times 10^{-2}\ \text{A}$$

or $$I_{\text{rms}} \approx 0.02396\ \text{A}.$$

The power actually dissipated (converted to heat) is only that in the resistor. The average power in the resistor is

$$P = I_{\text{rms}}^{2} R.$$

Substituting,

$$P = (0.02396)^2 \times 100\ \text{W} = 0.000574 \times 100\ \text{W} \approx 0.0574\ \text{W}.$$

The resistor has a heat capacity of $$2\ \text{J}/^{\circ}\text{C}$$. The energy needed to raise its temperature by $$\Delta T = 10^{\circ}\text{C}$$ is

$$Q = (\text{heat capacity}) \times \Delta T = 2 \times 10\ \text{J} = 20\ \text{J}.$$

If the whole of the electrical power goes into heating (no losses to the surroundings are assumed), the time required is

$$t = \dfrac{Q}{P} = \dfrac{20\ \text{J}}{0.0574\ \text{W}} \approx 3.47 \times 10^{2}\ \text{s} \approx 347\ \text{s}.$$

Rounded to the nearest whole second this is about $$348\ \text{s}$$.

Hence, the correct answer is Option D.