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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