A series LR circuit contains an emf source of 14 V with no internal resistance, a 34 H inductor with no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? 14 ohms 11 ohms 1.5 ohms 1.9 ohms 5.0 ohms
A series LR circuit contains an emf source of 14 V with no internal resistance, a 34 H inductor with no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4.0 s after the switch is closed, what is the resistance of the resistor? 14 ohms 11 ohms 1.5 ohms 1.9 ohms 5.0 ohms
The Correct Answer and Explanation is:
Correct Answer: 1.5 ohms
Explanation
In a series LR (inductor-resistor) circuit connected to a DC voltage source, the inductor initially opposes the change in current due to its property of self-inductance. Over time, the current increases and eventually reaches a maximum steady value, at which point the inductor acts like a short circuit (offering no opposition).
The key relationship for the emf across the inductor εL\varepsilon_L at time tt after the switch is closed is given by: εL(t)=ε0⋅e−Rt/L\varepsilon_L(t) = \varepsilon_0 \cdot e^{-Rt/L}
Where:
- εL(t)\varepsilon_L(t) is the emf across the inductor at time tt,
- ε0=initial emf=14 V\varepsilon_0 = \text{initial emf} = 14 \, \text{V},
- RR is the resistance in ohms,
- LL is the inductance in henries,
- tt is the time in seconds.
We’re told that 4.0 seconds after the switch is closed, the emf across the inductor is 80% of its maximum value, i.e., εL(4.0)=0.80⋅14=11.2 V\varepsilon_L(4.0) = 0.80 \cdot 14 = 11.2 \, \text{V}
Now substitute into the formula: 11.2=14⋅e−R⋅4/3411.2 = 14 \cdot e^{-R \cdot 4 / 34}
Divide both sides by 14: 0.80=e−4R/340.80 = e^{-4R/34}
Take the natural logarithm of both sides: ln(0.80)=−4R34\ln(0.80) = -\frac{4R}{34} −0.2231=−4R34-0.2231 = -\frac{4R}{34}
Multiply both sides by 34: −0.2231⋅34=−4R-0.2231 \cdot 34 = -4R −7.5854=−4R-7.5854 = -4R
Divide by -4: R=1.896≈1.9 ΩR = 1.896 \approx \boxed{1.9 \, \Omega}
However, note the error: We misinterpreted the decay formula. The voltage across the inductor is actually proportional to the rate of change of current, which decreases over time. The correct exponential behavior for the inductor voltage after closing the switch is: εL=ε⋅e−Rt/L\varepsilon_L = \varepsilon \cdot e^{-Rt/L}
So redoing the math properly: 0.80=e−Rt/L⇒ln(0.80)=−Rt/L⇒R=−Ltln(0.80)=−344ln(0.80)=8.5⋅0.2231≈1.90 Ω0.80 = e^{-Rt/L} \Rightarrow \ln(0.80) = -Rt/L \Rightarrow R = -\frac{L}{t} \ln(0.80) = -\frac{34}{4} \ln(0.80) = 8.5 \cdot 0.2231 \approx 1.90 \, \Omega
So the correct answer is: 1.9 ohms ✅
Final Answer: 1.9 ohms
