9.1 Electrical current  (Page 2/10)

Calculating the average current

The main purpose of a battery in a car or truck is to run the electric starter motor , which starts the engine. The operation of starting the vehicle requires a large current to be supplied by the battery. Once the engine starts, a device called an alternator takes over supplying the electric power required for running the vehicle and for charging the battery.

(a) What is the average current involved when a truck battery sets in motion 720 C of charge in 4.00 s while starting an engine? (b) How long does it take 1.00 C of charge to flow from the battery?

Strategy

We can use the definition of the average current in the equation $I=\frac{\text{Δ}Q}{\text{Δ}t}$ to find the average current in part (a), since charge and time are given. For part (b), once we know the average current, we can its definition $I=\frac{\text{Δ}Q}{\text{Δ}t}$ to find the time required for 1.00 C of charge to flow from the battery.

Solution

a. Entering the given values for charge and time into the definition of current gives

b. Solving the relationship $I=\frac{\text{Δ}Q}{\text{Δ}t}$ for time $\text{Δ}t$ and entering the known values for charge and current gives

Significance

a. This large value for current illustrates the fact that a large charge is moved in a small amount of time. The currents in these “starter motors” are fairly large to overcome the inertia of the engine. b. A high current requires a short time to supply a large amount of charge. This large current is needed to supply the large amount of energy needed to start the engine.

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Calculating instantaneous currents

Consider a charge moving through a cross-section of a wire where the charge is modeled as $Q\left(t\right)=Q_M\left(1-e^{\text{−}t\text{/}\tau }\right)$ . Here, $Q_M$ is the charge after a long period of time, as time approaches infinity, with units of coulombs, and $\tau$ is a time constant with units of seconds (see [link] ). What is the current through the wire?

Strategy

The current through the cross-section can be found from $I=\frac{dQ}{dt}$ . Notice from the figure that the charge increases to $Q_M$ and the derivative decreases, approaching zero, as time increases ( [link] ).

Solution

The derivative can be found using $\frac{d}{dx}{e}^u=e^u\frac{du}{dx}$ .

Significance

The current through the wire in question decreases exponentially, as shown in [link] . In later chapters, it will be shown that a time-dependent current appears when a capacitor charges or discharges through a resistor. Recall that a capacitor is a device that stores charge. You will learn about the resistor in Model of Conduction in Metals .

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