An AC source is connected to an inductance of $$100$$ mH, a capacitance of $$100$$ $$\mu$$F and a resistance of $$120$$ $$\Omega$$ as shown in figure. The time in which the resistance having a thermal capacity $$2$$ J °C$$^{-1}$$ will get heated by $$16°$$C is ______ s.

The solution to find the time required to heat the resistance in an LCR circuit involves calculating the circuit impedance, current, and the heat produced.

1. Calculation of Net Reactance and Impedance (Z)

First, we find the difference between inductive reactance ($$X_L$$) and capacitive reactance ($$X_C$$):

$$|X_L - X_C| = |10 - 100| = 90\ \Omega$$

The total impedance Z of the circuit is given by:

$$Z = \sqrt{(X_L - X_C)^2 + R^2}$$

Substituting the values ($$R = 120\ \Omega$$):

$$Z = \sqrt{(90)^2 + (120)^2} = \sqrt{8100 + 14400} = \sqrt{22500}$$

$$Z = 150\ \Omega$$

2. Calculation of RMS Current ($$I_{rms}$$)

Using Ohm's law for AC circuits ($$V_{rms} = 20\text{ V}$$ based on common context for this problem):

$$I_{rms} = \frac{V_{rms}}{Z} = \frac{20}{150} = \frac{2}{15}\text{ A}$$

3. Energy Balance and Time ($\Delta t$)

The electrical energy dissipated as heat in the resistor ($$I_{rms}^2 R \Delta t$$) is equal to the heat energy required to raise the temperature ($$ms\Delta T$$):

$$I_{rms}^2 R \Delta t = C_{thermal} (\Delta T)$$

Given:

• Thermal capacity (ms or $$C_{thermal}$$) $$= 2\text{ J °C}^{-1}$$
• Temperature change ($$\Delta T$$) = $$16\text{ °C}$$

Substituting the values:

$$\left( \frac{2}{15} \right)^2 \times 120 \times \Delta t = 2 \times 16$$

$$\frac{4}{225} \times 120 \times \Delta t = 32$$

$$\frac{480}{225} \Delta t = 32$$

$$\Delta t = \frac{32 \times 225}{480}$$

$$\Delta t = \frac{7200}{480}$$

$$\boxed{\Delta t = 15\text{ sec}}$$