An ac circuit contains a resistance of 1 ~kΩ$1 \mathrm{~k}\Omega$, a capacitor of 0.1 ~μ F$0.1 \mathrm{~\mu F}$ and an inductor of 1 ~mH$1 \mathrm{~mH}$ connected in series. The resonance frequency of the circuit is approximately.

Related Formula

ω = 2π f$\omega = 2\pi f$

Core Logic

Given: Peak current ipeak = 5 ~A$i_{\text{peak}} = 5 \mathrm{~A}$ Frequency f = 60 ~Hz$f = 60 \mathrm{~Hz}$ Angular frequency ω = 2π × 60 = 120π ~rad/s$\omega = 2\pi \times 60 = 120\pi \mathrm{~rad/s}$

The equation of the alternating current is:

We need to find the time t$t$ when the current i$i$ reaches its peak value (5 ~A$5 \mathrm{~A}$) for the first time.

Step 1: Calculate Time

The smallest positive angle for which sine is 1$1$ is (π)/(2)$\frac{\pi}{2}$:

Pattern Recognition

An AC wave starting from zero reaches its first peak exactly at a quarter of a full time period (T/4$T/4$). Since T = 1/f = 1/60 ~s$T = 1/f = 1/60 \mathrm{~s}$, the quarter period is 1 / (4 × 60) = 1/240 ~s$1 / (4 \times 60) = 1/240 \mathrm{~s}$.

Chapter Mix

Class 12 Physics: Alternating Current