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Chapter 25: Problem 61

The potential difference across the terminals of a battery is \(8.40 \mathrm{~V}\) when there is a current of \(1.50 \mathrm{~A}\) in the battery from the negative to the positive terminal. When the current is \(3.50 \mathrm{~A}\) in the reverse direction, the potential difference becomes \(10.20 \mathrm{~V}\). (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

Short Answer

Expert verified
The internal resistance of the battery is approximately \(0.4 \, \Omega\) and the emf of the battery is approximately \(9.0 \, V\).

Step by step solution

01

Setup the expressions for voltage

Use Ohm's law (\(V = IR\)) to set up an expression for the voltage across the battery in each case. When the current flows from negative to positive terminal (let's call this case 1), the voltage \(V1\) is equal to the emf (\(E\)) minus the product of the current (\(I1 = 1.50 A\)) and the internal resistance (\(r\)): \(V1 = E - I1 * r\). When the current flows in the reverse direction (case 2), the voltage \(V2\) is equal to the emf plus the product of the current (\(I2 = 3.50 A\)) and the internal resistance: \(V2 = E + I2 * r\).
02

Solve for internal resistance

You can solve for the internal resistance (\(r\)) by subtracting equation from Step 1, case 1 from equation from Step 1, case 2: \( V2 - V1= (E + I2 * r) - (E - I1 * r) = I2 * r + I1 * r = (I1 + I2) * r\). You can now solve for \(r\): \( r = (V2 - V1) / (I1 + I2)\). Plug into this formula the given values: \( r = (10.20 V - 8.40 V) / (1.50 A + 3.50 A) \).
03

Find the emf of the battery

Once you have found \(r\), you can substitute it back into either of the original equations from Step 1 to find the emf \(E\). Using the equation from case 1, \(E = V1 + I1 * r\), plug in the given and found values: \(E = 8.40 V + 1.50 A * r\).

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Key Concepts

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

Ohm's Law
Ohm's Law is one of the fundamental principles in electricity. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The formula given by Ohm's Law is:
  • \( V = IR \)
Where:
  • \( V \) is the voltage (in volts).
  • \( I \) is the current (in amperes).
  • \( R \) is the resistance (in ohms).
Ohm's Law helps us understand how voltage, current, and resistance interact in an electrical circuit. For example, if you increase the resistance while keeping the voltage constant, the current will decrease. This simple relationship is incredibly useful for solving problems in circuits and is often used to calculate unknown variables when other quantities are known. In our exercise, it's instrumental in setting up equations to solve for internal resistance and electromotive force.
Internal Resistance
Internal resistance is a property of all batteries and power supplies. It represents the resistance to the flow of electric current within the battery itself. When current flows through the battery, the internal resistance causes some of the energy to be lost as heat. This can affect the terminal voltage of the battery.

In formula terms, when you measure a lower terminal voltage than expected, it's likely due to the internal resistance. The modified version of Ohm's Law for a battery becomes:
  • \( V = E - Ir \)
Where:
  • \( E \) is the electromotive force (emf).
  • \( r \) is the internal resistance.
  • \( I \) is the current.
Our exercise used this concept to differentiate between the emf and actual terminal voltages for different current directions, enabling the computation of the internal resistance.
Electromotive Force (emf)
The electromotive force (emf) of a battery is the maximum potential difference the battery can provide. Think of it as the energy supplied by the battery to each coulomb of charge.

Emf is defined as the voltage across the terminals of the battery when no current is flowing. In real-life circumstances, when a current flows, the presence of internal resistance means that the terminal voltage is usually less than the emf.
  • \( E = V + Ir \)
Where:
  • \( E \) is the emf.
  • \( V \) is the terminal voltage.
  • \( I \) is the current flowing through the battery.
  • \( r \) is the internal resistance.
In the exercise scenario, understanding emf allowed us to determine the net energy delivered to the electric circuit when currents varied in magnitude and direction, affirming the relationship between the internal components of the battery.

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Most popular questions from this chapter

\(A 3.00 \mathrm{~m}\) length of copper wire at \(20^{\circ} \mathrm{C}\) has a \(1.20-\mathrm{m}\) -long section with diameter \(1.60 \mathrm{~mm}\) and a \(1.80-\mathrm{m}\) -long section with diameter \(0.80 \mathrm{~mm}\). There is a current of \(2.5 \mathrm{~mA}\) in the 1.60 -mm-diameter section. (a) What is the current in the 0.80 -mm-diameter section? (b) What is the magnitude of \(\vec{E}\) in the 1.60 -mm-diameter section? (c) What is the magnitude of \(\vec{E}\) in the 0.80 -mm-diameter section? (d) What is the potential difference between the ends of the \(3.00 \mathrm{~m}\) length of wire?

At temperatures near absolute zero, \(B_{\mathrm{c}}\) approaches \(0.142 \mathrm{~T}\) for vanadium, a type-I superconductor. The normal phase of vanadium has a magnetic susceptibility close to zero. Consider a long, thin vanadium cylinder with its axis parallel to an external magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) in the \(+x\) -direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectors are parallel to the \(x\) -axis. At temperatures near absolute zero, what are the resultant magnetic field \(\vec{B}\) and the magnetization \(\vec{M}\) inside and outside the cylinder (far from the ends) for (a) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.130 \mathrm{~T}) \hat{\boldsymbol{\imath}}\) and (b) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.260 \mathrm{~T}) \hat{\imath} ?\)

Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of \(3.00 \mathrm{~cm}^{2}\) and are separated by a 2.50 -mm- thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant \(4.70 .\) (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is \(120 \mathrm{~V}\) and the conduction current \(i_{\mathrm{C}}\) equals \(6.00 \mathrm{~mA} .\) At this instant, what are (a) the charge \(q\) on each plate; (b) the rate of change of charge on the plates; (c) the displacement current in the dielectric?

A \(25.0-\mathrm{cm}\) -long metal rod lies in the \(x y\) -plane and makes an angle of \(36.9^{\circ}\) with the positive \(x\) -axis and an angle of \(53.1^{\circ}\) with the positive \(y\) -axis. The rod is moving in the \(+x\) -direction with a speed of \(6.80 \mathrm{~m} / \mathrm{s}\). The rod is in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.120 \mathrm{~T}) \hat{\imath}-(0.220 \mathrm{~T}) \hat{\jmath}-(0.0900 \mathrm{~T}) \hat{\boldsymbol{k}} .\) (a) What is the magni- tude of the emf induced in the rod? (b) Indicate in a sketch which end of the rod is at higher potential.

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of \(165.0 \mathrm{~cm}\), but its circumference is decreasing at a constant rate of \(12.0 \mathrm{~cm} / \mathrm{s}\) due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude \(0.500 \mathrm{~T}\). (a) Find the emf induced in the loop at the instant when \(9.0 \mathrm{~s}\) have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
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