The value of current at resonance in a series L-C-R circuit is affected by the value of

Options

A. R only
B. C only
C. L only
D. L, C and R

Answer

To determine how the value of current at resonance in a series L-C-R circuit is affected, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Circuit**:
- We have a series L-C-R circuit connected to an AC source. The components are:
- L: Inductor
- C: Capacitor
- R: Resistor

2. **Impedance Calculation**:
- The impedance (Z) of the circuit is given by the formula:
\[ Z = \sqrt{(X_L - X_C)^2 + R^2} \]
- Where \(X_L\) is the inductive reactance and \(X_C\) is the capacitive reactance.

3. **Resonance Condition**:
- At resonance, the inductive reactance equals the capacitive reactance:
\[ X_L = X_C \]
- Therefore, at resonance, the impedance simplifies to:
\[ Z = R \]

4. **Current Calculation**:
- The current (I) in the circuit can be calculated using Ohm's law:
\[ I = \frac{E_{RMS}}{Z} \]
- Substituting the impedance at resonance:
\[ I = \frac{E_{RMS}}{R} \]

5. **Conclusion**:
- From the equation \(I = \frac{E_{RMS}}{R}\), we can see that the current at resonance is directly affected by the resistance (R) only. The values of L and C do not affect the current at resonance since they cancel each other out.

### Final Answer:
The value of current at resonance in a series L-C-R circuit is affected by the value of **R only**.

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