A series R - L - C `(R = 10 Omega, X_(L) = 20 Omega, X_(c) = 20 Omega)` circuit is supplied by V = 10 `sin omegat` volt then power dissipation in circuit is :-

Options

A. Zero
B. 10 watt
C. 5 watt
D. 2.5 watt

Answer

To solve the problem of power dissipation in a series R-L-C circuit, we will follow these steps:

### Step 1: Identify Given Values
We have the following values from the problem:
- Resistance, \( R = 10 \, \Omega \)
- Inductive reactance, \( X_L = 20 \, \Omega \)
- Capacitive reactance, \( X_C = 20 \, \Omega \)
- Voltage, \( V(t) = 10 \sin(\omega t) \)

### Step 2: Calculate the Impedance of the Circuit
In a series R-L-C circuit, the total impedance \( Z \) can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Since \( X_L = X_C \), we have:
\[ X_L - X_C = 20 - 20 = 0 \]
Thus, the impedance simplifies to:
\[ Z = \sqrt{R^2} = R = 10 \, \Omega \]

### Step 3: Calculate the Phase Angle \( \phi \)
The phase angle \( \phi \) can be found using:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
Substituting the values:
\[ \tan \phi = \frac{0}{10} = 0 \]
This implies:
\[ \phi = 0^\circ \]
Thus, \( \cos \phi = 1 \).

### Step 4: Calculate the RMS Voltage
The RMS voltage \( E_{rms} \) can be calculated from the peak voltage \( V_0 \):
\[ E_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, V \]

### Step 5: Calculate the Peak Current \( I_0 \)
The peak current \( I_0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{V_0}{Z} = \frac{10}{10} = 1 \, A \]

### Step 6: Calculate the RMS Current
The RMS current \( I_{rms} \) is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, A \]

### Step 7: Calculate Power Dissipation
The power dissipated in the circuit is given by the formula:
\[ P = I_{rms} \cdot E_{rms} \cdot \cos \phi \]
Substituting the values we calculated:
\[ P = \left(\frac{1}{\sqrt{2}}\right) \cdot \left(\frac{10}{\sqrt{2}}\right) \cdot 1 \]
\[ P = \frac{10}{2} = 5 \, W \]

### Final Answer
Thus, the power dissipation in the circuit is \( \boxed{5 \, W} \).

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