The electric resonance is sharp in L-C-R circuit if in the circuit -

Options

A. R is greater
B. R is smaller
C. R = `X_(L) "or" X_(C)`
D. Does not depend on R

Answer

To determine when the electric resonance is sharp in an L-C-R circuit, we need to analyze the relationship between resistance (R), inductive reactance (X_L), and capacitive reactance (X_C). Here’s a step-by-step solution:

### Step 1: Understand Resonance Condition
In an L-C-R circuit, resonance occurs when the inductive reactance (X_L) equals the capacitive reactance (X_C). This can be expressed mathematically as:
\[ X_L = X_C \]

### Step 2: Write the Expression for Impedance
The total impedance (Z) of the L-C-R circuit can be expressed as:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
At resonance, since \( X_L = X_C \), the equation simplifies to:
\[ Z = R \]

### Step 3: Analyze Current at Resonance
The current (I) in the circuit can be expressed as:
\[ I = \frac{E_0 \sin(\omega t)}{Z} \]
At resonance, substituting Z gives:
\[ I = \frac{E_0 \sin(\omega t)}{R} \]

### Step 4: Determine the Effect of Resistance on Current
From the equation \( I = \frac{E_0 \sin(\omega t)}{R} \), we can see that:
- As R decreases, the current I increases.
- This means that for a sharper resonance peak, we want R to be as small as possible.

### Step 5: Conclusion
Therefore, the electric resonance is sharp in an L-C-R circuit when the resistance R is minimized. This leads us to conclude that the correct answer to the question is:
**The electric resonance is sharp in L-C-R circuit if R is smaller.**

### Final Answer:
**Option B: R is smaller.**

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