MHTCET 20 Apr 2026 Shift 1 Paper

Concept:In a series $LCR$ circuit, the r.m.s. current is maximum at resonance and decreases on both sides of the resonant frequency.Explanation:For a series $LCR$ circuit, the impedance is$$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{R^{2}+\left(2\pi fL-\frac{1}{2\pi fC}\right)^{2}}$$The r.m.s. current is$$i=\frac{V_{\text{rms}}}{Z}$$So, when impedance $Z$ is minimum, current $i$ is maximum.At resonance,$$f_{r}=\frac{1}{2\pi\sqrt{LC}}$$At this frequency, $X_{L}=X_{C}$, hence $\left(X_{L}-X_{C}\right)=0$.Thus, $Z$ becomes minimum,$$Z_{\min}=R$$Therefore, current reaches its maximum value at $f=f_{r}$.For $f < f_{r}$, impedance increases, so current decreases.For $f>f_{r}$, impedance increases again, so current decreases.Hence, the $i$ versus $f$ graph rises to a sharp peak at resonance and then falls.This variation is correctly shown by graph $(P)$.Answer:Option B: $(P)$