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A battery has emf \(30.0 \mathrm{~V}\) and internal resistance \(r .\) A \(9.00 \Omega\) resistor is connected to the terminals of the battery, and the voltage drop across the resistor is \(27.0 \mathrm{~V}\). What is the internal resistance of the battery?

Short Answer

Expert verified
The internal resistance of the battery is 1.0 Ohm.

Step by step solution

01

Understand the given

The given values are emf (E) = 30.0V, resistance (R) = 9.0Ω, voltage drop across the resistor (V) = 27.0 V. The unknown value we require is internal resistance (r).
02

Apply formulas

According to Ohm's law, total terminal voltage (V) is equal to current (I) times the total resistance. Where total resistance includes both internal resistance and the resistance from resistor i.e., Total Resistance = r + R. But we already know the current can be calculated as I = V/R. Hence, writing Eq (1) in terms of I gives E = I * (R + r).
03

Calculate current

Using the given values for V and R, calculate the current. I = V/R = 27.0V/9.0 Ohm = 3.0 A.
04

Calculate internal resistance

Inserting the obtained value of current and given values into Eq (1), we solve for r. E = I * (R + r). So, 30.0V = 3.0A * (9.0 Ohm + r), simplifying we get, r = 30.0V/3.0A - 9.00 Ohm = 1.0 Ohm.

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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 a fundamental principle in the field of electrical engineering and physics, which describes the relationship between voltage, current, and resistance in an electrical circuit. It can be expressed using the simple formula:
\( V = I \times R \),
where \( V \) is the voltage across the resistance, \( I \) is the current flowing through the resistance, and \( R \) is the resistance.

Applying Ohm's Law can help solve various problems related to electrical circuits, such as finding the value of current given voltage and resistance, or determining the resistance necessary to achieve a certain current flow with a given voltage. For example, if you have a resistor with a known resistance and you measure the voltage drop across it, you can calculate the current using this law. This concept was crucial in the step-by-step solution provided, as it allowed us to calculate the current (I) in the circuit with the known voltage drop (V) and resistor's resistance (R).
Electromotive Force (emf)
Electromotive force, or emf, is not actually a force, but rather a measure of the energy provided by a source (such as a battery or generator) per charge to move around a circuit. It's often thought of as the 'pressure' that pushes charges through a conducting loop, much like water pressure pushes water through a pipe. The emf of a power source can be greater than the voltage measured across its terminals due to the internal resistance of the device.

For our problem, the emf represents the maximum potential difference the battery could provide if there was no internal resistance to impede the flow of the current. The emf often gets confused with terminal voltage, but unlike terminal voltage, emf considers both the internal and external circuit. In the solution, emf helps us to calculate the internal resistance \( r \) of the battery by using the formula \( E = I \times (R + r) \), where \( E \) is the emf.
Electrical Circuits
Electrical circuits are pathways that allow electric current to flow through them and are composed of various electrical components, such as resistors, capacitors, and inductors, among others. The understanding of electrical circuits is critical as they form the basis for electronic devices and systems that we use in everyday life.

In simple terms, a circuit can be series or parallel, and each configuration has its own unique impact on the total resistance, current distribution, and voltage drops across the components. In our context, we've tackled a series circuit that includes a battery with internal resistance and an external resistor. The total voltage supplied by the battery must account for the voltage drop across both the internal resistance and the external load resistance. This concept is vital for understanding how different components of a circuit affect the overall electrical properties and behavior of the circuit, as well as for calculating the internal resistance of the battery from the voltage drops observed.

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Most popular questions from this chapter

A motor vehicle generates electrical power using an alternator, which employs electromagnetic induction to convert mechanical energy to electrical energy. The alternator acts as a dc generator (Example 29.4 ). The alternator maintains and replenishes charge on the car's battery and operates headlights, radiator fans, windshield wipers, power windows, computer systems, sensors, sound systems, and other components. (a) A typical car battery provides 70 amp-hours of charge. How many coulombs is that? (b) If headlights each draw 20 A of current, a radiator fan draws \(10 \mathrm{~A},\) and windshield wipers each draw \(5 \mathrm{~A},\) estimate the peak current needed for a car to operate on a rainy night. (c) A car's alternator supplies an average emf of \(14 \mathrm{~V}\) as emf induced in a sequence of stator coils in the presence of a magnetic field created by rotor coil electromagnets turned by a pulley system. A stator coil may have 42 windings and a cross-sectional diameter of \(5.0 \mathrm{~cm},\) and it rotates at \(400 \mathrm{~Hz}\). Estimate the strength of the magnetic field generated by a rotor coil.

The armature of a small generator consists of a flat, square coil with 120 turns and sides with a length of \(1.60 \mathrm{~cm} .\) The coil rotates in a magnetic field of \(0.0750 \mathrm{~T}\). What is the angular speed of the coil if the maximum emf produced is \(24.0 \mathrm{mV} ?\)

A ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

A very long, straight solenoid with a cross-sectional area of \(2.00 \mathrm{~cm}^{2}\) is wound with 90.0 turns of wire per centimeter. Starting at \(t=0\) the current in the solenoid is increasing according to \(i(t)=\left(0.160 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2}\). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is \(3.20 \mathrm{~A}\) ?

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