# finding angular frequency when current is maximum in circuit

A series of ac circuit consist of an inductor and capacitor. The inductance and capacitance are 1 henry and 25μf. The current is maximum in circuit, then what will be the angular frequency?

The answer is 200. Could somone explain how???

The current in a series resonant circuit will be at a maximum when:

$$f = \frac{1}{2\pi \sqrt{LC}} \text{ ,}$$

so we can solve for that frequency like this:

$$f = \frac{1}{6.28\times \sqrt{\text{1H} \times 2.5 \cdot10^{-5}\text{F}}} = 31.85\text {Hz}$$

and then the angular frequency will be:

$$\omega o = 2\pi \text{f} = 6.28\times 31.85\text{Hz} \approx \text {200 radians per second}$$

Or, simply:

$$\omega o = \frac{1}{ \sqrt{LC}} \text{ radians per second }$$

You will get the maximum current when the impedance is minimum. The impedance of this simple series circuit is:

$$Z = j \cdot \omega \cdot L - j \cdot \dfrac{1}{\omega \cdot C} = j \left( \omega \cdot L - \cdot \dfrac{1}{\omega \cdot C} \right)$$

Which we see we can make zero when

$$\omega \cdot L = \dfrac{1}{\omega \cdot C} \Rightarrow \omega = \dfrac{1}{\sqrt{L \cdot C}}$$

Taking $L = 1 \text{ henry}$, $C = 25 \mu F = 25 \times 10^{-6} \text{ Farad}$

$$\omega = \dfrac{1}{\sqrt{1 \cdot 25 \times 10^{-6}}} = 200 \text{ rad per sec}$$

The current is maximum when impedance is zero. Now the value for frequency when impedance is zero, also known as resonant frequency is equal to inverse to sqroot of Inductance and capacitance. It should be $\frac{1000}{\sqrt{25}} = 200$.

• thank you for the support. I found a small typo on question and fixed now. Still u think the answer is wrong? – ABHILASH Jul 19 '15 at 7:20