Question

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

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

Step-by-step solution

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.

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: $$0.204 = \frac{6}{R}(X - e^{\frac{-0.01}{(L/R)}})$$

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: $$0.273 = \frac{6}{R}\left(\frac{0.204}{(6/R)} + e^{\frac{-0.01}{(L/R)}}\right)$$

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: $$R = \frac{60}{\ln\left(\frac{0.273}{0.204} - 1\right)} \approx 116.99 \Omega$$

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 \(\approx 50.65 \mathrm{mH}\).