10.5 Rc circuits  (Page 3/5)

Strategy

Solution

Significance

RC circuits have many applications. They can be used effectively as timers for applications such as intermittent windshield wipers, pace makers, and strobe lights. Some models of intermittent windshield wipers use a variable resistor to adjust the interval between sweeps of the wiper. Increasing the resistance increases the RC time constant, which increases the time between the operation of the wipers.

Another application is the pacemaker . The heart rate is normally controlled by electrical signals, which cause the muscles of the heart to contract and pump blood. When the heart rhythm is abnormal (the heartbeat is too high or too low), pace makers can be used to correct this abnormality. Pacemakers have sensors that detect body motion and breathing to increase the heart rate during physical activities, thus meeting the increased need for blood and oxygen, and an RC timing circuit can be used to control the time between voltage signals to the heart.

Looking ahead to the study of ac circuits ( Alternating-Current Circuits ), ac voltages vary as sine functions with specific frequencies. Periodic variations in voltage, or electric signals, are often recorded by scientists. These voltage signals could come from music recorded by a microphone or atmospheric data collected by radar. Occasionally, these signals can contain unwanted frequencies known as “noise.” RC filters can be used to filter out the unwanted frequencies.

In the study of electronics, a popular device known as a 555 timer provides timed voltage pulses. The time between pulses is controlled by an RC circuit. These are just a few of the countless applications of RC circuits.

Intermittent windshield wipers

A relaxation oscillator is used to control a pair of windshield wipers. The relaxation oscillator consists of a 10.00-mF capacitor and a $10.00\text{-k}\text{Ω}$ variable resistor known as a rheostat. A knob connected to the variable resistor allows the resistance to be adjusted from $0.00\text{Ω}$ to $10.00\text{k}\text{Ω}.$ The output of the capacitor is used to control a voltage-controlled switch. The switch is normally open, but when the output voltage reaches 10.00 V, the switch closes, energizing an electric motor and discharging the capacitor. The motor causes the windshield wipers to sweep once across the windshield and the capacitor begins to charge again. To what resistance should the rheostat be adjusted for the period of the wiper blades be 10.00 seconds?

Strategy

The resistance considers the equation $V_{\text{out}}\left(t\right)=V\left(1-e^{\text{−}t\text{/}\tau }\right),$ where $\tau =RC.$ The capacitance, output voltage, and voltage of the battery are given. We need to solve this equation for the resistance.

Solution

The output voltage will be 10.00 V and the voltage of the battery is 12.00 V. The capacitance is given as 10.00 mF. Solving for the resistance yields

$\begin{array}{}\\ \\ \hfill V_{\text{out}}\left(t\right)& =\hfill & V\left(1-e^{\text{−}t\text{/}\tau }\right),\hfill \\ \hfill e^{\text{−}t\text{/}RC}& =\hfill & 1-\frac{V_{\text{out}}\left(t\right)}{V},\hfill \\ \hfill \text{ln}\left(e^{\text{−}t\text{/}RC}\right)& =\hfill & \text{ln}\left(1-\frac{V_{\text{out}}\left(t\right)}{V}\right),\hfill \\ \hfill -\frac{t}{RC}& =\hfill & \text{ln}\left(1-\frac{V_{\text{out}}\left(t\right)}{V}\right),\hfill \\ \hfill R& =\hfill & \frac{\text{−}t}{C\text{ln}\left(1-\frac{V_C\left(t\right)}{V}\right)}=\frac{\mathrm{-10.00}\text{s}}{10\times {10}^{\mathrm{-3}}\text{F}\text{ln}\left(1-\frac{10\text{V}}{12\text{V}}\right)}=558.11\text{Ω}.\hfill \end{array}$

Significance

Increasing the resistance increases the time delay between operations of the windshield wipers. When the resistance is zero, the windshield wipers run continuously. At the maximum resistance, the period of the operation of the wipers is:

$t=\text{−}RC\text{ln}\left(1-\frac{V_{\text{out}}\left(t\right)}{V}\right)=\text{−}\left(10\times {10}^{\mathrm{-3}}\text{F}\right)\left(10\times {10}^3\text{Ω}\right)\text{ln}\left(1-\frac{10\text{V}}{12\text{V}}\right)=179.18\text{s}=2.98\text{min}.$

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