Effective value of current in a circuit connected across an AC source at 60 Hz is 8A . Find the value of current `5xx10^(-3)` s after it was zero .

Answer

To solve the problem, we need to find the current at a specific time after it was zero in an AC circuit. Here are the steps to arrive at the solution:

### Step 1: Determine the Peak Current (I₀)
The effective value of current (I_RMS) is given as 8 A. We can find the peak current (I₀) using the formula:

\[ I_{RMS} = \frac{I_0}{\sqrt{2}} \]

Rearranging this formula gives us:

\[ I_0 = I_{RMS} \times \sqrt{2} \]

Substituting the given value:

\[ I_0 = 8 \, \text{A} \times \sqrt{2} \approx 8 \times 1.414 = 11.31 \, \text{A} \]

### Step 2: Write the Expression for Current as a Function of Time
The current in an AC circuit can be expressed as:

\[ I(t) = I_0 \sin(\omega t) \]

where \(\omega\) (angular frequency) is given by:

\[ \omega = 2\pi f \]

For a frequency (f) of 60 Hz:

\[ \omega = 2\pi \times 60 \approx 376.99 \, \text{rad/s} \]

### Step 3: Calculate the Time Value
We need to find the current at \(t = 5 \times 10^{-3} \, \text{s}\):

\[ I(t) = I_0 \sin(\omega t) \]

Substituting the values we have:

\[ I(5 \times 10^{-3}) = 11.31 \sin(376.99 \times 5 \times 10^{-3}) \]

Calculating the argument of the sine function:

\[ \omega t = 376.99 \times 0.005 \approx 1.88495 \, \text{rad} \]

### Step 4: Calculate the Sine Value
Now, we find:

\[ I(5 \times 10^{-3}) = 11.31 \sin(1.88495) \]

Using a calculator to find \(\sin(1.88495)\):

\[ \sin(1.88495) \approx 0.951 \]

### Step 5: Calculate the Current
Now substituting back into the equation:

\[ I(5 \times 10^{-3}) \approx 11.31 \times 0.951 \approx 10.76 \, \text{A} \]

### Final Answer
The value of the current \(5 \times 10^{-3}\) seconds after it was zero is approximately **10.76 A**.

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