For a series L-C-R circuit with L=1.00 mH, C=10`muF` and `R=50Omega`, is driven with 5V AC voltage. At resosance, the current through the circuit is

Options

A. 0.2 A
B. 0.25 A
C. 0.15 A
D. 0.1 A

Explanation

To find the current through a series L-C-R circuit at resonance, we can follow these steps:

### Step 1: Identify the given values
- Inductance (L) = 1.00 mH = 1.00 x 10^-3 H
- Capacitance (C) = 10 µF = 10 x 10^-6 F
- Resistance (R) = 50 Ω
- Voltage (V) = 5 V

### Step 2: Determine the resonance condition
At resonance in a series L-C-R circuit, the impedance (Z) is equal to the resistance (R). Therefore, we can write:
\[ Z = R \]

### Step 3: Calculate the current at resonance
The current (I) through the circuit at resonance can be calculated using Ohm's Law:
\[ I = \frac{V}{Z} \]
Since at resonance \( Z = R \), we can substitute:
\[ I = \frac{V}{R} \]

### Step 4: Substitute the values into the equation
Now, substituting the known values into the equation:
\[ I = \frac{5 \, \text{V}}{50 \, \Omega} \]

### Step 5: Perform the calculation
Calculating the current:
\[ I = \frac{5}{50} = 0.1 \, \text{A} \]

### Conclusion
The current through the circuit at resonance is:
\[ I = 0.1 \, \text{A} \]

### Final Answer
The correct option is D: 0.1 A.

---