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Chapter 27: Problem 22

The current in a series \(R L\) circuit increases to \(20 \%\) of its final value in \(3.1 \mu \mathrm{s}\). If \(L=1.8 \mathrm{mH},\) what's the resistance?

Short Answer

Expert verified
The resistance in the given series RL circuit is approximately \(249.76 \Omega\).

Step by step solution

01

Convert quantities to standard units

Convert the time \(t\) from microseconds to seconds so it's in standard SI units. This gives \(t = 3.1 \times 10^{-6} s\). Convert inductance \(L\) from millihenries to henries, giving \(L = 1.8 \times 10^{-3} H\).
02

Solve the time constant equation for \(R\)

The formula for the time constant in an RL circuit is \(\tau = L/R\). We want to solve for \(R\) so rewrite the equation as \(R = L/\tau\). We'll find \(\tau\) in the next step and then substitute it and \(L\) into this equation to find \(R\).
03

Substitute values to find \(\tau\)

The equation \(i = I_{f}(1 - e^{-\frac{t}{\tau}})\) can be rearranged when \(I_{f} = 1\) and \(i = 0.2\) to give \(\tau = -\frac{t}{\ln (1 - i)}\). By substituting the given values into this equation, we find that \(\tau = -\frac{3.1 \times 10^{-6} s}{\ln (1 - 0.2)} = 7.207 \times 10^{-6} s\).
04

Substitute values to find \(R\)

Now substitute the values of \(\tau\) and \(L\) from steps 1 and 3 into the equation \(R = L/\tau\), giving \(R = \frac{1.8 \times 10^{-3} H}{7.207 \times 10^{-6} s}\). The result is \(R = 249.76 \Omega\).

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Most popular questions from this chapter

You and your roommate are headed to Cancún for spring break. Your roommate, who has had only high school physics, has read that an emf can be induced in the wings of an airplane and wonders whether this would give enough voltage to power a portable music player. What's your answer? (Assume that the wingspan of your 747 is \(60 \mathrm{m},\) the plane is flying at 600 mph, and Earth's magnetic field is 0.3 G.)

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