A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?
The Correct Answer and Explanation is:
To find the current through an RL (resistor-inductor) circuit after a given time, we use the following formula for the current in an RL circuit connected to a DC source: I(t)=VR(1−e−t/τ)I(t) = \frac{V}{R} \left(1 – e^{-t/\tau}\right)
Where:
- I(t)I(t) is the current at time tt,
- VV is the voltage of the power supply (15 V),
- RR is the resistance (60 Ω),
- LL is the inductance (45 mH = 0.045 H),
- τ=LR\tau = \frac{L}{R} is the time constant of the RL circuit,
- ee is Euler’s number (~2.718),
- tt is the time since the switch was closed (7.0 ms = 0.007 s).
Step 1: Calculate the time constant τ\tau
τ=LR=0.04560=0.00075 s=0.75 ms\tau = \frac{L}{R} = \frac{0.045}{60} = 0.00075 \, \text{s} = 0.75 \, \text{ms}
Step 2: Plug into the formula
I(t)=1560(1−e−0.007/0.00075)I(t) = \frac{15}{60} \left(1 – e^{-0.007/0.00075} \right) I(t)=0.25(1−e−9.33)I(t) = 0.25 \left(1 – e^{-9.33} \right)
Since e−9.33≈8.84×10−5e^{-9.33} \approx 8.84 \times 10^{-5}, we get: I(t)=0.25(1−8.84×10−5)I(t) = 0.25 \left(1 – 8.84 \times 10^{-5} \right) I(t)≈0.25×0.9999116≈0.249978 AI(t) \approx 0.25 \times 0.9999116 \approx 0.249978 \, \text{A}
✅ Final Answer:
\boxed{0.250 \, \text{A}} \] (rounded to three significant figures) — ### Explanation (300 words): When an inductor is suddenly connected in series with a resistor and a DC power source, the current does not immediately jump to its final value. This is due to the inductor’s property of opposing changes in current. The opposition arises from the self-induced EMF generated when the current through the inductor changes. The inductor essentially “resists” the change, causing the current to rise gradually, not instantly. The rate at which current increases is governed by the **time constant** \( \tau = \frac{L}{R} \), which represents the time it takes for the current to reach about 63.2% of its maximum value. After several time constants, the current gets very close to its steady-state value \( I = \frac{V}{R} \). In this case, with a 45-mH inductor and 60-Ω resistor, the time constant is 0.75 ms. At 7 ms (roughly 9.33 time constants), the current has had plenty of time to rise close to its maximum value. Mathematically, at this point, the exponential decay term \( e^{-t/\tau} \) becomes nearly zero. Therefore, the current approaches: \[ I = \frac{V}{R} = \frac{15}{60} = 0.25 \, \text{A}
This shows that by 7 ms, the current has effectively reached steady state, and the inductor is now acting like a short circuit, allowing full current to flow as though only the resistor were in the circuit.
