A 45-mH ideal inductor is connected in series with a 60-Q 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 ms later?
250 mA
650 mA
500 mA
800 mA
The Correct Answer and Explanation is :
To solve for the current in the circuit 7 ms after closing the switch, we need to analyze the behavior of the RL circuit. The equation governing the current in an RL circuit is:
[
I(t) = \frac{V}{R} \left( 1 – e^{-\frac{R}{L} t} \right)
]
Where:
- ( I(t) ) is the current at time ( t ),
- ( V ) is the voltage of the power supply (15 V),
- ( R ) is the resistance (60 Ω),
- ( L ) is the inductance (45 mH = 0.045 H),
- ( t ) is the time (7 ms = 0.007 s).
Step 1: Calculate the time constant
The time constant ( \tau ) of an RL circuit is given by:
[
\tau = \frac{L}{R}
]
Substituting the given values:
[
\tau = \frac{0.045}{60} = 0.00075 \, \text{s} \, (0.75 \, \text{ms})
]
Step 2: Use the formula for current
Now, we can calculate the current at ( t = 7 \, \text{ms} ) (which is 0.007 s) using the equation for current in an RL circuit:
[
I(t) = \frac{V}{R} \left( 1 – e^{-\frac{R}{L} t} \right)
]
Substitute the known values:
[
I(0.007) = \frac{15}{60} \left( 1 – e^{-\frac{60}{0.045} \times 0.007} \right)
]
First, calculate the exponent:
[
\frac{60}{0.045} = 1333.33, \quad 1333.33 \times 0.007 = 9.3333
]
So,
[
I(0.007) = 0.25 \left( 1 – e^{-9.3333} \right)
]
Since ( e^{-9.3333} ) is a very small value close to 0, the expression simplifies to:
[
I(0.007) \approx 0.25 \times (1 – 0) = 0.25 \, \text{A} = 250 \, \text{mA}
]
Final Answer:
The current 7 ms after the switch is closed is 250 mA.