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To solve for the value of angular frequency \( \omega \) such that the peak current \( i(t) \) is 2 A, let's analyze the circuit.

### Step 1: Circuit Analysis
This is an AC circuit with a resistor and capacitor in series. The voltage source is given as:
\[
v(t) = 25 \cos(\omega t) \, \text{V}
\]
The impedance \( Z \) of a series RC circuit is given by:
\[
Z = R + \frac{1}{j \omega C}
\]
where:
- \( R = 10 \, \Omega \)
- \( C = 0.002 \, \text{F} \)
- \( j \) is the imaginary unit.

The magnitude of the impedance is:
\[
|Z| = \sqrt{R^2 + \left( \frac{1}{\omega C} \right)^2}
\]

### Step 2: Relating Voltage and Current
The peak current \( I_{\text{max}} \) in the circuit is given by:
\[
I_{\text{max}} = \frac{V_{\text{max}}}{|Z|}
\]
where \( V_{\text{max}} = 25 \) V. We are told that \( I_{\text{max}} = 2 \) A, so:
\[
2 = \frac{25}{\sqrt{10^2 + \left( \frac{1}{\omega \times 0.002} \right)^2}}
\]

### Step 3: Solving for \( \omega \)
Rearrange this equation to solve for \( \omega \):
\[
\sqrt{10^2 + \left( \frac{1}{\omega \times 0.002} \right)^2} = \frac{25}{2} = 12.5
\]

Square both sides to remove the square root:
\[
10^2 + \left( \frac{1}{\omega \times 0.002} \right)^2 = 12.5^2
\]
\[
100 + \frac{1}{(0.002 \omega)^2} = 156.25
\]

Isolate \( \omega \):
\[
\frac{1}{(0.002 \omega)^2} = 156.25 - 100
\]
\[
\frac{1}{(0.002 \omega)^2} = 56.25
\]
\[
(0.002 \omega)^2 = \frac{1}{56.25}
\]
\[
0.002 \omega = \frac{1}{\sqrt{56.25}}
\]
\[
0.002 \omega = \frac{1}{7.5}
\]
\[
\omega = \frac{1}{7.5 \times 0.002}
\]
\[
\omega = \frac{1}{0.015} = 66.67 \, \text{rad/s}
\]

### Answer
The required value of \( \omega \) is approximately:
\[
\omega \approx 66.67 \, \text{rad/s}
\]

Would you like more details on any step, or do you have further questions?

### Related Questions:
1. How does the impedance of a capacitor change with frequency?
2. What would happen to the current if the resistance were doubled?
3. How is the phase angle between voltage and current determined in an RC circuit?
4. What is the significance of the angular frequency \( \omega \) in AC circuits?
5. How does a change in capacitance affect the resonance frequency of an RC circuit?

### Tip:
In AC circuits, the current depends on the impedance, which varies with frequency due to capacitors and inductors.

Determine the value of ω in the circuit shown below such that the peak (maximum) value of current i(t) is 2A.

Learn how to determine the value of angular frequency (ω) in a series RC circuit such that the peak current i(t) is 2A. This detailed solution covers AC circuit analysis, impedance calculation, and step-by-step equations. Suitable for undergraduate students studying electrical engineering.

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