A series LR circuit contains an emf source of 14 V having no internal resistance, a resistor, a 34 H inductor having no appreciable resistance, and a switch. If the em across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? A) 1.9 Ω B) 1.50 Ω C) 14 Ω D) 5.0 Ω
The Correct Answer and Explanation is :
To solve this problem, we need to analyze the behavior of an LR circuit when the switch is closed, which initiates the current flow from the emf source (in this case, 14 V) through the resistor (R) and inductor (L).
Circuit Analysis
When the switch is closed, the current ( I(t) ) in an LR circuit increases according to the formula:
[
I(t) = \frac{E}{R} \left(1 – e^{-\frac{R}{L} t}\right)
]
where:
- ( E ) is the emf (14 V)
- ( R ) is the resistance (unknown)
- ( L ) is the inductance (34 H)
- ( t ) is the time (4.0 s)
The maximum current ( I_{\text{max}} ) in the circuit can be calculated using:
[
I_{\text{max}} = \frac{E}{R}
]
The voltage across the inductor ( V_L(t) ) is given by:
[
V_L(t) = L \frac{dI}{dt}
]
Using the derivative of ( I(t) ), we have:
[
\frac{dI}{dt} = \frac{E}{L} e^{-\frac{R}{L} t}
]
Thus,
[
V_L(t) = L \cdot \frac{E}{L} e^{-\frac{R}{L} t} = E e^{-\frac{R}{L} t}
]
At maximum current, ( V_L ) approaches zero, and the voltage across the inductor is maximum when the current is at its maximum.
Given Conditions
It is given that ( V_L(4) = 0.8 \cdot V_{L, \text{max}} ).
Calculating the maximum voltage across the inductor when current is maximum:
[
V_{L, \text{max}} = E = 14 \text{ V}
]
So,
[
V_L(4) = 0.8 \cdot 14 = 11.2 \text{ V}
]
Setting Up the Equation
Using the expression for ( V_L(t) ):
[
11.2 = 14 \cdot e^{-\frac{R}{34} \cdot 4}
]
Solving for ( R ):
[
e^{-\frac{R}{34} \cdot 4} = \frac{11.2}{14} \Rightarrow e^{-\frac{R}{34} \cdot 4} \approx 0.8
]
Taking the natural logarithm of both sides:
[
-\frac{R}{34} \cdot 4 = \ln(0.8)
]
Calculating ( \ln(0.8) \approx -0.2231 ):
[
-\frac{R \cdot 4}{34} = -0.2231 \Rightarrow \frac{R \cdot 4}{34} = 0.2231
]
Solving for ( R ):
[
R = \frac{0.2231 \cdot 34}{4} \approx 1.90 \, \Omega
]
Conclusion
The resistance of the resistor in the circuit is approximately 1.9 Ω.
Therefore, the correct answer is A) 1.9 Ω.