A 1.9 H inductor, a `100 mu F` capacitor and `25 Omega` resistor are connected in series to an a.c. source whose e.m.f. (in volt) varies with time t (in seconds) according to the expression `E = 282 sin 100t`. Determine (i) the reactance, (ii) the impedance (iii) the r.m.s. value of the current and (iv) the rate of dissipation of heat

Answer

To solve the problem step by step, we will determine the reactance, impedance, r.m.s. value of the current, and the rate of dissipation of heat in the given RLC circuit.

### Given Data:
- Inductance, \( L = 1.9 \, \text{H} \)
- Capacitance, \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Resistance, \( R = 25 \, \Omega \)
- E.M.F. of the source, \( E = 282 \sin(100t) \)

### Step 1: Determine the Angular Frequency (\( \omega \))
From the given E.M.F. equation \( E = 282 \sin(100t) \), we can identify that:
\[ \omega = 100 \, \text{rad/s} \]

### Step 2: Calculate Inductive Reactance (\( X_L \))
The inductive reactance is given by:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 100 \times 1.9 = 190 \, \Omega \]

### Step 3: Calculate Capacitive Reactance (\( X_C \))
The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{100 \times 100 \times 10^{-6}} = \frac{1}{0.01} = 100 \, \Omega \]

### Step 4: Calculate the Impedance (\( Z \))
The impedance of the series RLC circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting the values:
\[ Z = \sqrt{25^2 + (190 - 100)^2} = \sqrt{625 + 90^2} = \sqrt{625 + 8100} = \sqrt{8725} \approx 93.407 \, \Omega \]

### Step 5: Calculate the r.m.s. Value of the Current (\( I_{\text{rms}} \))
The r.m.s. voltage (\( V_{\text{rms}} \)) is given by:
\[ V_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{282}{\sqrt{2}} \approx 199.79 \, \text{V} \]
The r.m.s. current is then calculated using:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
Substituting the values:
\[ I_{\text{rms}} = \frac{199.79}{93.407} \approx 2.134 \, \text{A} \]

### Step 6: Calculate the Rate of Dissipation of Heat (Power)
The power dissipated in the circuit is given by:
\[ P = I_{\text{rms}}^2 R \]
Substituting the values:
\[ P = (2.134)^2 \times 25 \approx 11.39 \times 25 \approx 284.75 \, \text{W} \]

### Summary of Results:
1. Inductive Reactance, \( X_L = 190 \, \Omega \)
2. Capacitive Reactance, \( X_C = 100 \, \Omega \)
3. Impedance, \( Z \approx 93.407 \, \Omega \)
4. r.m.s. Current, \( I_{\text{rms}} \approx 2.134 \, \text{A} \)
5. Rate of Dissipation of Heat, \( P \approx 284.75 \, \text{W} \)