A series RL circuit has $V = 10\text{ V}$, $R = 5 \ \Omega$, $L = 2\text{ H}$. With $i(t) = (V/R)(1 - e^{-t/\tau})$, $\tau = L/R$. Find $i$ at $t = 0.5\text{ s}$.
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For quick calculations in exams, remember that at \(t = 1\tau\), current reaches \(63.2\%\) of steady-state value.
At \(t = 2\tau\), it reaches \(86.5\%\).
Since \(t = 0.5\text{ s}\) is slightly more than \(1\tau = 0.4\text{ s}\), the current must be slightly more than \(63.2\%\) of \(2\text{ A}\) (\(1.264\text{ A}\)).
Among the options, \(1.428\text{ A}\) is the only reasonable value that is greater than \(1.264\text{ A}\) but less than \(2\text{ A}\).
Step 1: Understanding the Problem:
The question asks to find the current \(i(t)\) flowing through a series RL circuit at time \(t = 0.5\text{ s}\), given the circuit parameters: voltage \(V = 10\text{ V}\), resistance \(R = 5\ \Omega\), and inductance \(L = 2\text{ H}\). Step 2: Key Formula or Approach:
The transient current in a series RL circuit is given by:
\[ i(t) = \frac{V}{R}\left(1 - e^{-t/\tau}\right) \]
where the time constant \(\tau\) is:
\[ \tau = \frac{L}{R} \]
Step 3: Detailed Explanation:
• First, calculate the time constant \(\tau\) using the given values \(L = 2\text{ H}\) and \(R = 5\ \Omega\):
\[ \tau = \frac{2}{5} = 0.4\text{ s} \]
• Determine the steady-state current value \(I_{\text{ss}} = \frac{V}{R}\):
\[ I_{\text{ss}} = \frac{10}{5} = 2\text{ A} \]
• Write down the expression for current \(i(t)\):
\[ i(t) = 2\left(1 - e^{-t/0.4}\right) = 2\left(1 - e^{-2.5t}\right) \]
• Substitute \(t = 0.5\text{ s}\) into the equation:
\[ i(0.5) = 2\left(1 - e^{-2.5 \times 0.5}\right) \]
\[ i(0.5) = 2\left(1 - e^{-1.25}\right) \]
• Compute the value of the exponential term:
\[ e^{-1.25} \approx 0.2865 \]
• Substitute this back into the current equation:
\[ i(0.5) = 2 \times (1 - 0.2865) = 2 \times 0.7135 = 1.427\text{ A} \]
• This is approximately \(1.428\text{ A}\), which matches Option (B). Step 4: Final Answer:
The current at \(t = 0.5\text{ s}\) is \(1.428\text{ A}\).
