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A wire with resistance \(R\) is connected to the terminals of a \(6.0 \mathrm{V}\) battery. What is the potential difference \(\Delta V_{\text {ends }}\) between the ends of the wire and the current \(I\) through it if the wire has the following resistances? (a) \(1.0 \Omega\) (b) \(2.0 \Omega\) (c) \(3.0 \Omega.\)

Short Answer

Expert verified
(a) \(\Delta V_{ends} = 6.0 V\), \(I = 6.0 A\); (b) \(\Delta V_{ends} = 6.0 V\), \(I = 3.0 A\); (c) \(\Delta V_{ends} = 6.0 V\), \(I = 2.0 A\).

Step by step solution

01

Determine the potential difference

The potential difference across the ends of the wire is equal to the voltage of the battery because the wire is directly connected to the battery. Therefore, \(\Delta V_{ends} = 6.0 V\).
02

Calculate Current for \(R = 1.0 \, \Omega\)

Ohm's law can be rearranged to solve for current: \(I = \Delta V / R\). Substituting \(\Delta V = 6.0 V\) and \(R = 1.0 \, \Omega\), we get \(I = 6.0 V / 1.0 \, \Omega = 6.0 A\).
03

Calculate Current for \(R = 2.0 \, \Omega\)

Substituting \(\Delta V = 6.0 V\) and \(R = 2.0 \, \Omega\) into the rearranged Ohm's law, we get \(I = 6.0 V / 2.0 \, \Omega = 3.0 A\).
04

Calculate Current for \(R = 3.0 \, \Omega\)

Substituting \(\Delta V = 6.0 V\) and \(R = 3.0 \, \Omega\) into the rearranged Ohm's law, we get \(I = 6.0 V / 3.0 \, \Omega = 2.0 A\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Potential Difference (Voltage)
Understanding electric potential difference, commonly known as voltage, is crucial when studying electrical circuits. Voltage is the driving force that pushes electric charges through a conductor and is measured in volts (V). It's analogous to the pressure difference that moves water through pipes. In our textbook example, the wire connected to a battery experiences a potential difference, simply because one terminal of the battery is at a higher electric potential than the other.

To visualize this concept, imagine a battery as a pump that creates pressure (voltage) to move charges through the wire—an essential movement that we call electric current. In the exercise, with the battery providing a steady 6.0 V, the potential difference across the wire's ends remains consistent at \(\Delta V_{\text{ends}} = 6.0 V\) no matter the resistance of the wire. This example illustrates a fundamental property of electric circuits: as long as there's a continuous path, the voltage supplied by the source will appear across the circuit.
Electric Current
Electric current is the rate at which charge is flowing through a conductor. Measured in amperes (A), current tells us how much charge passes a point in the circuit per second. Think of it as the volume of water flowing through a hose—the larger the volume, the higher the flow. In electrical terms, a high current means many charges are moving past a point each second.

In the given exercise, the current varies depending on the resistance of the wire. Ohm's Law, which defines the relationship between voltage, current, and resistance, is given by \(I = \Delta V / R\). This means that when the resistance increases (for instance, going from 1.0 \(\rOmega:\Omega\) to 3.0 \(\rOmega:\Omega\)), the current decreases, as seen with the calculated currents: 6.0 A, 3.0 A, and 2.0 A, respectively. This inversely proportional relationship between current and resistance helps us predict how changes in resistance will affect the current flow.
Electrical Resistance
Electrical resistance is the property of a material that opposes the flow of electric current. It dictates how much current will flow for a given voltage. Resistance is like the friction that water experiences in a pipe—tighter and longer pipes increase resistance, just as thinner wires or those made of less conductive materials increase electrical resistance.

In our textbook problem, we observe how different resistance values affect the flow of current through the wire when connected to a 6.0 V battery. The calculations show us a simple relationship: as resistance increases, if the voltage remains the same, the current decreases. This is again described by Ohm's Law, \(R = \Delta V / I\), which can be rearranged to calculate the current for a given voltage and resistance. When we plug in the different resistance values (1.0 \(\rOmega:\Omega\), 2.0 \(\rOmega:\Omega\), 3.0 \(\rOmega:\Omega\)), we get corresponding currents that clearly demonstrate how resistance affects the flow of electric current.

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