Question:easy

Which graph shows the correct variation of r.m.s. current '\(i\)' with frequency '\(f\)' of a.c. in case of series resonant circuit ?

Show Hint

Current rises to a maximum at resonance and falls on both sides, so the graph is a smooth peak.
Updated On: Oct 1, 2026
  • \((R)\)
  • \((P)\)
  • \((S)\)
  • \((Q)\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Think about the limits:
For $f \to 0$, the capacitor blocks the current, so $i \to 0$. For $f \to \infty$, the inductor blocks it, so $i \to 0$.

Step 2: In between:
There must be a frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$ where the two reactances cancel and the current is largest. The graph is therefore a single-peaked bell curve.

Step 3: Match:
Graph (P) is the smooth bell. Graphs (Q) and (S) have a dip, and (R) has a kink-like spike.

Final Answer:
Graph (P), option (B). \[ \boxed{\text{(P)}} \]
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