A series `R-L-C` circuit has `R = 100` ohm. `L = 0.2 mH` and `C = (1)/(2) mu F`. The applied voltage `V = 20 sin omega t`. Then
At resonant frequency `omega_(0), ((V_(R ))_("max"))/((V_(L))_("max")) =`

Options

A. 2
B. 5
C. 3
D. 4

Answer

To solve the problem, we need to find the ratio of the maximum voltage across the resistor \( V_R \) to the maximum voltage across the inductor \( V_L \) at the resonant frequency \( \omega_0 \) in a series RLC circuit.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Resistance \( R = 100 \, \Omega \)
- Inductance \( L = 0.2 \, \text{mH} = 0.2 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = \frac{1}{2} \, \mu F = 0.5 \times 10^{-6} \, \text{F} \)
- Voltage \( V = 20 \sin(\omega t) \)

2. **Calculate the Resonant Frequency \( \omega_0 \):**
The resonant frequency for a series RLC circuit is given by:
\[ \omega_0 = \frac{1}{\sqrt{LC}} \]
Substituting the values:
\[ \omega_0 = \frac{1}{\sqrt{(0.2 \times 10^{-3})(0.5 \times 10^{-6})}} \]
\[ = \frac{1}{\sqrt{0.1 \times 10^{-9}}} = \frac{1}{10^{-5}} = 10^5 \, \text{rad/s} \]

3. **Calculate the Maximum Voltage Across the Resistor \( V_R \):**
At resonance, the maximum voltage across the resistor is given by:
\[ V_R = I_{max} \cdot R \]
Where \( I_{max} \) is the maximum current. The maximum current can be calculated using the maximum voltage \( V_0 \):
\[ I_{max} = \frac{V_0}{Z} \]
At resonance, the impedance \( Z \) is equal to \( R \):
\[ I_{max} = \frac{20}{100} = 0.2 \, \text{A} \]
Therefore,
\[ V_R = 0.2 \cdot 100 = 20 \, \text{V} \]

4. **Calculate the Maximum Voltage Across the Inductor \( V_L \):**
The maximum voltage across the inductor at resonance is given by:
\[ V_L = I_{max} \cdot X_L \]
Where \( X_L = \omega_0 L \):
\[ X_L = (10^5)(0.2 \times 10^{-3}) = 20 \, \Omega \]
Thus,
\[ V_L = 0.2 \cdot 20 = 4 \, \text{V} \]

5. **Calculate the Ratio \( \frac{V_R}{V_L} \):**
\[ \frac{V_R}{V_L} = \frac{20}{4} = 5 \]

### Final Answer:
\[ \frac{(V_R)_{max}}{(V_L)_{max}} = 5 \]