91Ó°ÊÓ

When dealing with electric circuits, one of the fundamental concepts involves the flow of current through a resistor. Current, measured in amperes (A), represents the flow of electric charge through a conductor, such as a wire or a resistor. In a circuit, the resistor is an element that opposes the flow of electric current, converting some of its energy into heat.

Ohm's Law, a key principle in electric circuits, states that current (\(I\)) through a conductor is directly proportional to the voltage (\(V\)) across it, and inversely proportional to the resistance (\(R\)) of the conductor. This relationship is defined by the equation:

• \(I = \frac{V}{R}\)

Using this formula lets us determine how much current will flow through a resistor if the voltage across it is known. This is essential in circuit design and analysis, as it helps predict the behavior of electrical components.

Voltage in many electrical circuits is often provided in the form of a sine wave. This alternating current (AC) type of voltage changes periodically, following the shape of a sine curve. In mathematical terms, sine wave voltage can be represented as:

• \(v(t) = V_m \sin(\omega t)\)

Here,

• \(V_m\) is the peak voltage or the maximum value that the voltage reaches.
• \(\omega\) represents angular frequency and determines how fast the sine wave oscillates per second. It's related to the standard frequency (\(f\)) by the equation:\(\omega = 2\pi f\).

In a circuit problem such as our exercise, identifying the sine wave voltage function correctly is crucial since it helps us understand how the voltage varies over time.

• This, in turn, permits us to calculate other variables, like the current through a resistor, using Ohm's Law.

Resistance is a measure of how much a resistor opposes the current flowing through it. In the exercise, you might have noticed we started with a known resistance value of \(50\Omega\). Understanding how to work with resistance is vital since it's a key component in calculating the current.

Once the voltage function and resistance are known, calculating the current becomes a matter of applying Ohm's Law. For example, substituting our given values into the equation:

• Voltage: \((160V) \sin(120\pi t)\)
• Resistance: \(50\Omega\)

We get:

• \(I(t) = \frac{(160V) \sin(120\pi t)}{50\Omega}\)
• \(I(t) = (3.2A) \sin(120\pi t)\)

Breaking down complex parts into simpler components makes these calculations manageable, ensuring accuracy in solving circuit problems. It's important in homework and exams to master this process for success.