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Chapter 18: Problem 84

A charging \(R C\) circuit controls the intermittent windshield wipers in a car. The emf is \(12.0 \mathrm{V}\). The wipers are triggered when the voltage across the 125 - \(\mu\) F capacitor reaches \(10.0 \mathrm{V} ;\) then the capacitor is quickly discharged (through a much smaller resistor) and the cycle repeats. What resistance should be used in the charging circuit if the wipers are to operate once every \(1.80 \mathrm{s} ?\)

Short Answer

Expert verified
Answer: The required resistance for the charging circuit is approximately \(1143.72 \, \Omega\).

Step by step solution

01

Identify the variables in the problem

The given values are: Voltage when the capacitor in the charging circuit is triggered (V_c): 10.0V Initial voltage across the capacitor (V_0): 12.0V Capacity of the capacitor (C): 125µF Time it takes for the wiper to operate (t): 1.80s Resistance (R): Unknown
02

Re-write the formula in terms of R

We are trying to find R, so we need to rewrite the equation in terms of R. Since we have: \(V_c(t) = V_0(1 - e^{-t/(RC)})\) We can divide both sides by V_0 to isolate the exponential part: \(\frac{V_c}{V_0} = 1 - e^{-t/(RC)}\) We can now make it easier to solve for R: \(e^{-t/(RC)} = 1 - \frac{V_c}{V_0}\) \(-t/(RC) = \ln(1 - \frac{V_c}{V_0})\) \(R = -t / (C \ln(1 - \frac{V_c}{V_0}))\)
03

Plug in the values and solve for R

Now, we can plug in the values to find R: \(R = -1.80 / (125 \times 10^{-6} \ln(1 - \frac{10.0}{12.0}))\) Calculate the value of R: \(R \approx 1143.72 \, \Omega\) Therefore, a resistance of approximately \(1143.72 \, \Omega\) should be used in the charging circuit for the windshield wipers to operate once every 1.80 seconds.

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Most popular questions from this chapter

An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?
A wire with cross-sectional area \(A\) carries a current \(I\) Show that the electric field strength \(E\) in the wire is proportional to the current per unit area \((I / A)\) and identify the constant of proportionality. [Hint: Assume a length \(L\) of wire. How is the potential difference across the wire related to the electric field in the wire? (Which is uniform?) Use \(V=I R\) and the connection between resistance and resistivity.]
A 20 - \(\mu\) F capacitor is discharged through a 5 -k\Omega resistor. The initial charge on the capacitor is \(200 \mu \mathrm{C}\) (a) Sketch a graph of the current through the resistor as a function of time. Label both axes with numbers and units. (b) What is the initial power dissipated in the resistor? (c) What is the total energy dissipated?
A portable CD player does not have a power rating listed, but it has a label stating that it draws a maximum current of \(250.0 \mathrm{mA}\). The player uses three \(1.50-\mathrm{V}\) batteries connected in series. What is the maximum power consumed?
Two copper wires, one double the diameter of the other, have the same current flowing through them. If the thinner wire has a drift speed \(v_{1},\) and the thicker wire has a drift speed \(v_{2},\) how do the drift speeds of the charge carriers compare?
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