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Chapter 20: Problem 59

A coil of wire is connected to an ideal \(6.00-\mathrm{V}\) battery at \(t=0 .\) At \(t=10.0 \mathrm{ms},\) the current in the coil is \(204 \mathrm{mA}\) One minute later, the current is 273 mA. Find the resistance and inductance of the coil. [Hint: Sketch \(I(t) .]\)

Short Answer

Expert verified
Answer: The resistance of the coil is approximately 116.99 Ω and the inductance is approximately 50.65 mH.

Step by step solution

01

1. Determine the equation for the current through the coil with time

We know that the current through an inductor (L) can be described using the equation: $$I(t) = \frac{V}{R}(1 - e^{\frac{-t}{(L/R)}})$$ Where \(I(t)\) is the current at time t, V is the voltage across the inductor, R is the resistance, and L is the inductance.
02

2. Calculate the time constant (τ) using given values

Given, at \(t = 0.01s\), \(I(t) = 0.204A\) and at \(t = 60.01s\), \(I(t) = 0.273A\). First, we need to solve for τ (the time constant). We can use the second condition to find τ: $$0.273 = \frac{6}{R}(1 - e^{\frac{-60.01}{(L/R)}})$$ Let's call \((1 - e^{\frac{-60.01}{(L/R)}})\) as X. So, $$0.273 = \frac{6}{R}X$$ By using the first condition, we can form the equation: $$0.204 = \frac{6}{R}(X - e^{\frac{-0.01}{(L/R)}})$$
03

3. Calculate X from the first condition

Now, let's isolate X in the first condition: $$X = \frac{0.204}{(6/R)} + e^{\frac{-0.01}{(L/R)}}$$ Substitute this expression of X into the second condition: $$0.273 = \frac{6}{R}\left(\frac{0.204}{(6/R)} + e^{\frac{-0.01}{(L/R)}}\right)$$
04

4. Solve for R using the second condition

Now, let's solve for R: $$\frac{0.273}{0.204} = \left(1 + \frac{e^{\frac{-0.01}{(L/R)}}}{e^{\frac{-60.01}{(L/R)}}}\right)$$ From the above equation, we can deduce that: $$\frac{0.273}{0.204} - 1 = \frac{e^{\frac{-0.01}{(L/R)}}}{e^{\frac{-60.01}{(L/R)}}} = e^{\frac{60}{(L/R)}}$$ Taking natural logarithm on both sides, we get: $$\ln\left(\frac{0.273}{0.204} - 1\right) = \frac{60}{(L/R)}$$ Knowing V = 6 volts, we can calculate the resistance R: $$R = \frac{60}{\ln\left(\frac{0.273}{0.204} - 1\right)} \approx 116.99 \Omega$$
05

5. Calculate L using the value of R and X

Now that we have R, we can find L using the value of X from the first condition: $$X = \frac{0.204}{(6/116.99)} + e^{\frac{-0.01}{(L/116.99)}}$$ $$X = \frac{0.204}{(6/116.99)} = 3.936$$ Now ∴ $$3.936 = 1 - e^{\frac{-60.01}{(L/116.99)}}$$ Solving for L, we get: $$L = \frac{116.99}{60.01}\ln\left(\frac{1}{1-3.936}\right) \approx 50.65 \mathrm{mH}$$ So, the resistance of the coil is \(\approx 116.99 \Omega\) and the inductance is \(\approx 50.65 \mathrm{mH}\).

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Most popular questions from this chapter

In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
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