A series LR circuit contains an emf source of 14V having no internal resistance, a resistor, a 34H inductor having 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, calculate the resistance of the resistor.
The Correct Answer and Explanation is:
Given:
- emf E=14 VE = 14 \, V (source voltage)
- Inductance L=34 HL = 34 \, H
- Time t=4.0 st = 4.0 \, s
- Voltage across inductor at t=4.0 st = 4.0 \, s is 80% of its maximum value
- No internal resistance in the emf source and negligible resistance in the inductor
- Need to find resistance RR
Step 1: Understand the circuit behavior
When the switch is closed in an LRLR series circuit, the current builds up according to: i(t)=ER(1−e−RLt)i(t) = \frac{E}{R} \left(1 – e^{-\frac{R}{L}t}\right)
The voltage across the inductor is: VL(t)=LdidtV_L(t) = L \frac{di}{dt}
Taking derivative of i(t)i(t): didt=ER⋅RLe−RLt=ELe−RLt\frac{di}{dt} = \frac{E}{R} \cdot \frac{R}{L} e^{-\frac{R}{L}t} = \frac{E}{L} e^{-\frac{R}{L}t}
Therefore: VL(t)=L×ELe−RLt=Ee−RLtV_L(t) = L \times \frac{E}{L} e^{-\frac{R}{L}t} = E e^{-\frac{R}{L}t}
Step 2: Voltage across the inductor at t=4t = 4 s
- The maximum voltage across the inductor just after closing the switch (at t=0t=0) is equal to the emf E=14 VE = 14 \, V.
- At t=4 st=4 \, s, the voltage across the inductor is 80% of its maximum:
VL(4)=0.8×14=11.2 VV_L(4) = 0.8 \times 14 = 11.2 \, V
From the equation above: VL(t)=Ee−RLtV_L(t) = E e^{-\frac{R}{L} t}
So, 11.2=14e−R34×411.2 = 14 e^{-\frac{R}{34} \times 4}
Divide both sides by 14: 0.8=e−4R340.8 = e^{-\frac{4R}{34}}
Take natural logarithm: ln(0.8)=−4R34\ln(0.8) = -\frac{4R}{34} −0.2231=−4R34-0.2231 = -\frac{4R}{34}
Multiply both sides by −1-1: 0.2231=4R340.2231 = \frac{4R}{34}
Solve for RR: R=0.2231×344=7.58544=1.896 ΩR = \frac{0.2231 \times 34}{4} = \frac{7.5854}{4} = 1.896 \, \Omega
Final answer:
R≈1.9 Ω\boxed{R \approx 1.9 \, \Omega}
Explanation:
When an LR circuit is energized by closing the switch, the current through the circuit increases gradually from zero toward its maximum value determined by Ohm’s law, Imax=ERI_{max} = \frac{E}{R}. The inductor initially opposes the sudden change in current, and the voltage across it is maximum (equal to the emf) at the instant the switch is closed.
Over time, as the current increases, the voltage across the inductor decreases exponentially, following the relation: VL=Ee−RLtV_L = E e^{-\frac{R}{L}t}
This exponential decay is governed by the time constant τ=LR\tau = \frac{L}{R}. The time constant indicates how quickly the current reaches its steady-state value and how fast the inductor’s voltage decays.
Given that at 4 seconds the inductor’s voltage is 80% of its initial value, we use this exponential decay to find the resistance. By taking the natural logarithm of the voltage ratio, we can solve for RR.
Physically, a higher resistance RR causes the current to rise more slowly and the voltage across the inductor to decay more quickly, because the time constant τ\tau becomes smaller. Conversely, a smaller RR means a slower current increase and slower decay in VLV_L.
In this problem, the resistance is approximately 1.9 Ω1.9\, \Omega, which balances the inductance and time given the voltage decay observed.
