When V = 100 `sinomegat` is applied across a series (R-L-C) circuit, At resonance the current in resistance `(R=100 `ohm`)` is i = `i_(0)` `sinomegat`, then power dissipation in circuit is:-

Options

A. 50 W
B. 100 W
C. 25 W
D. Can't be calculated

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The voltage applied across the circuit is \( V = 100 \sin(\omega t) \).
- The resistance \( R = 100 \, \Omega \).

2. **Determine the Maximum Voltage (\( V_0 \)):**
- From the voltage equation, we can identify that the maximum voltage \( V_0 = 100 \).

3. **Calculate the RMS Voltage (\( V_{RMS} \)):**
- The RMS voltage is given by:
\[ V_{RMS} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2} \, V \]

4. **Determine the Current at Resonance:**
- At resonance, the current is given as \( i = i_0 \sin(\omega t) \).
- The maximum current \( i_0 \) can be calculated using Ohm's law:
\[ i_0 = \frac{V_0}{R} = \frac{100}{100} = 1 \, A \]

5. **Calculate the RMS Current (\( I_{RMS} \)):**
- The RMS current is given by:
\[ I_{RMS} = \frac{i_0}{\sqrt{2}} = \frac{1}{\sqrt{2}} \, A \]

6. **Determine the Power Factor (\( \cos \phi \)):**
- At resonance, the phase difference \( \phi = 0 \), hence:
\[ \cos \phi = 1 \]

7. **Calculate the Power Dissipation (\( P \)):**
- The power dissipated in the circuit can be calculated using the formula:
\[ P = V_{RMS} \times I_{RMS} \times \cos \phi \]
- Substituting the values:
\[ P = (50\sqrt{2}) \times \left(\frac{1}{\sqrt{2}}\right) \times 1 = 50 \, W \]

### Final Answer:
The power dissipation in the circuit is \( P = 50 \, W \).

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