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?
a.250 ma
b.850 ma
c.550 ma
d.280 ma
e.650 ma
The Correct Answer and Explanation is :
The correct answer is: 250 mA
To find the current through the circuit 7.0 ms after the switch is closed, we can use the formula for the current in an RL (resistor-inductor) circuit:
[
I(t) = I_{\text{max}} \left(1 – e^{-\frac{R}{L} t}\right)
]
where:
- ( I(t) ) is the current at time ( t ),
- ( I_{\text{max}} ) is the maximum steady-state current,
- ( R ) is the resistance,
- ( L ) is the inductance,
- ( t ) is the time in seconds,
- ( e ) is the base of the natural logarithm (approximately equal to 2.71828).
Step 1: Calculate ( I_{\text{max}} )
The maximum current ( I_{\text{max}} ) can be calculated using Ohm’s law:
[
I_{\text{max}} = \frac{V}{R}
]
where:
- ( V = 15 \, \text{V} ) (the voltage of the power supply),
- ( R = 60 \, \Omega ).
Substituting the values:
[
I_{\text{max}} = \frac{15 \, \text{V}}{60 \, \Omega} = 0.25 \, \text{A} \text{ or } 250 \, \text{mA}
]
Step 2: Find the time constant ( \tau )
The time constant ( \tau ) of the RL circuit is given by:
[
\tau = \frac{L}{R}
]
Substituting the given values:
[
L = 45 \, \text{mH} = 0.045 \, \text{H}
]
[
R = 60 \, \Omega
]
[
\tau = \frac{0.045}{60} = 0.00075 \, \text{s} = 0.75 \, \text{ms}
]
Step 3: Calculate the current at ( t = 7.0 \, \text{ms} )
Now, we can substitute ( t = 7.0 \, \text{ms} = 0.007 \, \text{s} ) into the current formula:
[
I(0.007) = 0.25 \left(1 – e^{-\frac{60}{0.045} \cdot 0.007}\right)
]
Calculating ( \frac{R}{L} \cdot t ):
[
\frac{60}{0.045} \cdot 0.007 \approx 9.333
]
Now substituting back into the equation:
[
I(0.007) = 0.25 \left(1 – e^{-9.333}\right) \approx 0.25 \left(1 – 0.000083\right) \approx 0.25 \times 0.999917 \approx 0.24998 \, \text{A}
]
This means ( I(0.007) \approx 0.25 \, \text{A} = 250 \, \text{mA} ).
Conclusion
Thus, the current 7.0 ms after closing the switch is approximately 250 mA, which corresponds to option a.