Question 9 10 point A 45 mH ideal inductor is connected in series with a 60- resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t=0s, what is the current 7.0 ms later?
A) 250 mA
B) 650 mA
C) 550 mA
D) 280 mA
E) 850 mA
The Correct Answer and Explanation is :
To solve this problem, we need to analyze the behavior of an RL circuit after the switch is closed at time ( t = 0 ). The given parameters are:
- Inductor ( L = 45 \, \text{mH} = 0.045 \, \text{H} )
- Resistor ( R = 60 \, \Omega )
- Voltage supply ( V = 15 \, \text{V} )
- Time ( t = 7.0 \, \text{ms} = 0.007 \, \text{s} )
Step 1: The RL circuit equation
When the switch is closed at ( t = 0 ), the current in an RL circuit increases gradually according to the following equation:
[
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 applied voltage,
- ( R ) is the resistance,
- ( L ) is the inductance,
- ( t ) is the time after the switch is closed.
Step 2: Calculate the time constant
The time constant ( \tau ) of an RL circuit is given by:
[
\tau = \frac{L}{R}
]
Substitute the given values:
[
\tau = \frac{0.045}{60} = 0.00075 \, \text{s}
]
So, the time constant is ( 0.75 \, \text{ms} ).
Step 3: Find the current at ( t = 7.0 \, \text{ms} )
Now, substitute the known values into the current equation to find the current at ( t = 0.007 \, \text{s} ):
[
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} \times 0.007 = 9.33
]
So, the equation becomes:
[
I(0.007) = 0.25 \left( 1 – e^{-9.33} \right)
]
Since ( e^{-9.33} ) is a very small number, we approximate:
[
I(0.007) \approx 0.25 \times (1 – 0) = 0.25 \, \text{A} = 250 \, \text{mA}
]
Final Answer:
The current 7.0 ms after the switch is closed is approximately 250 mA. The correct answer is:
A) 250 mA.