At t = 0, switch is closed, and switch 2 is closed 4 s later.
Find i(t) for t > 0. Calculate i for t = 2 s and t = 5 s.

To solve the problem of finding the current \( i(t) \) for \( t > 0 \) in a circuit where a switch is closed at \( t = 0 \) and another switch is closed 4 seconds later, we need to analyze the circuit behavior over time. This typically involves understanding the circuit components, such as resistors, capacitors, and inductors, and how they respond to changes in the circuit configuration.

Understanding the Circuit Behavior

When the first switch is closed at \( t = 0 \), the circuit begins to operate based on the initial conditions set by the components present. The second switch, which closes at \( t = 4 \) seconds, will alter the circuit's configuration, affecting the current flow.

Initial Conditions and Time Constants

Assuming we have a simple RC (resistor-capacitor) circuit, the current \( i(t) \) can be described by the following equations:

• For \( 0 < t < 4 \): The circuit behaves according to the first switch being closed.
• For \( t \geq 4 \): The circuit behavior changes due to the closing of the second switch.

Current Calculation for \( 0 < t < 4 \)

In the time interval from \( 0 \) to \( 4 \) seconds, the current can be expressed as:

i(t) = I_0 (1 - e^{-t/RC})

Here, \( I_0 \) is the initial current, \( R \) is the resistance, and \( C \) is the capacitance. The term \( e^{-t/RC} \) represents the exponential decay of the current over time.

Current Calculation for \( t \geq 4 \)

Once the second switch is closed at \( t = 4 \) seconds, the circuit configuration changes, and we need to recalculate the current. The new equation might look like:

i(t) = I_f (1 - e^{-(t-4)/RC})

In this case, \( I_f \) is the new steady-state current after the second switch closes.

Specific Current Values

Now, let's calculate the current at specific times:

Finding \( i(2) \)

For \( t = 2 \) seconds, we use the first equation:

i(2) = I_0 (1 - e^{-2/RC})

Substituting the known values of \( I_0 \), \( R \), and \( C \) will give us the current at this time.

Finding \( i(5) \)

For \( t = 5 \) seconds, we switch to the second equation since \( t \) is now greater than 4:

i(5) = I_f (1 - e^{-(5-4)/RC})

Again, substituting the known values will yield the current at this time.

Conclusion

By analyzing the circuit behavior before and after the second switch closes, we can effectively determine the current at any given time. Remember to substitute the actual values for \( R \), \( C \), and the initial or final currents to get numerical results for \( i(2) \) and \( i(5) \).