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Light Bulbs. The power rating of a light bulb (such as a \(100 \mathrm{~W}\) bulb is the power it dissipates when connected across a \(120 \mathrm{~V}\) potential difference. What is the resistance of (a) a \(100 \mathrm{~W}\) bulb and (b) a \(60 \mathrm{~W}\) bulb? (c) How much current does each bulb draw in normal use?

Short Answer

Expert verified
The resistance of a 100W light bulb is \(144 \, Ohms\) and of a 60W bulb is \(240 \, Ohms\). The current each bulb draws in normal use for 100W and 60W bulbs are \(0.83 \, A\) and \(0.50 \, A\), respectively.

Step by step solution

01

Calculate the resistance

The formula for power is \(P = V^2/R\) where \(P\) is power, \(V\) is voltage, and \(R\) is resistance. We need to rearrange this formula to calculate resistance: \(R = V^2/P\). we substitute \(V = 120 V\) and \(P = 100W\) (for the 100W bulb) and \(P = 60W\) (for the 60W bulb) into the rearranged power formula.
02

Calculate the current

To calculate the current drawn by each bulb, we use Ohm's Law given by \(I = V/R\), where \(I\) is current, \(V\) is voltage and \(R\) is resistance. We substitute \(V = 120 V\) and the calculated resistance \(R\) (for both bulbs separately) into the formula to calculate the current.

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Key Concepts

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

Resistance
Resistance is a fundamental concept in electric circuits. It refers to the opposition a material offers to the flow of electric current. Think of it as hurdles for electrons as they move through a conductor.

The unit of Resistance is the ohm (symbolized as \(\Omega\)). When we talk about resistance, we're often referring to Ohm's Law, but let's focus on how it acts in practical applications like a light bulb.

When you have a light bulb with a specific power rating, like 100 watts, this means it's designed to dissipate 100 watts of power when connected to a certain voltage, such as 120 volts in our example. The resistance of the bulb is a crucial factor, determining how much electricity the bulb will consume. You can calculate the resistance of a bulb using the formula derived from power equations:
  • For a 100W bulb: \( R = \frac{V^2}{P} = \frac{120^2}{100} \Omega \)
  • For a 60W bulb: \( R = \frac{120^2}{60} \Omega \)
By manipulating this formula, you can identify the resistance specific to any power-rated bulb in terms of volts and watts.
Ohm's Law
Ohm's Law is a cornerstone of electrical engineering and physics. It helps us comprehend how voltage, current, and resistance interrelate. The law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance: \( V = IR \).

But why does this matter? Ohm’s Law allows us to calculate the current, resistance, or voltage if the other two quantities are known. For example, in the context of a light bulb:
  • If you know the voltage (120 V) and have calculated the resistance from the power rating, you can easily find the current. Use the rearranged formula \( I = \frac{V}{R} \).
This not only gives insights into how much electricity a device uses, but also helps in designing circuits effectively to ensure they're safe and functional.
Current
Current is the flow of electric charge. It's like the river of electricity that flows through your circuit. It's one of the key components to understand how circuits work.

Current is measured in amperes (A), often referred to as "amps." By using Ohm's Law, you can determine how much current a specific device will draw when it's operational.

For instance, the current drawn by our hypothetical incandescent bulbs:
  • For a 100W bulb with a 120V voltage, first determine resistance using the formula \( R = \frac{120^2}{100} \) and then calculate \( I = \frac{120}{R} \).
  • Similarly, for a 60W bulb, determine its resistance and follow by substituting into the same current formula.
Thus, by understanding how much current flows through the bulb, not only do you ensure that the circuit breaker does not trip, but you can also evaluate the efficiency of electricity consumption in your setup.

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Most popular questions from this chapter

A material with resistivity \(\rho\) is formed into a cylinder of length \(L\) and outer radius \(r_{\text {outer }}\). A cylindrical core with radius \(r_{\text {inner }}\) is removed from the axis of this cylinder and filled with a conducting material, which is attached to a wire. The outer surface of the cylinder is coated with a conducting material and attached to another wire. (a) If the second wire has potential \(V\) greater than the first wire, in what direction does the local electric field point inside of the cylinder? (b) The magnitude of this electric field is \(c / r,\) where \(c\) is a constant and \(r\) is the distance from the axis of the cylinder. Use the relationship \(V=\int \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{l}}\) to determine the constant \(c .(\mathrm{c})\) What is the resistance of this device? (d) A \(1.00-\mathrm{cm}\) -long hollow cylindrical resistor has an inner radius of \(1.50 \mathrm{~mm}\) and an outer radius of \(3.00 \mathrm{~mm} .\) The material is a blend of powdered carbon and ceramic whose resistivity \(\rho\) may be altered by changing the amount of carbon. If this device should have a resistance of \(6.80 \mathrm{k} \Omega,\) what value of \(\rho\) should be selected?

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.

An electrical conductor designed to carry large currents has a circular cross section \(2.50 \mathrm{~mm}\) in diameter and is \(14.0 \mathrm{~m}\) long. The resistance between its ends is \(0.104 \Omega\). (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is \(1.28 \mathrm{~V} / \mathrm{m},\) what is the total current? (c) If the material has \(8.5 \times 10^{28}\) free electrons per cubic meter, find the average drift speed under the conditions of part (b).

An incandescent light bulb uses a coiled filament of tungsten that is \(580 \mathrm{~mm}\) long with a diameter of \(46.0 \mu \mathrm{m} .\) At \(20.0^{\circ} \mathrm{C}\) tungsten has a resistivity of \(5.25 \times 10^{-8} \Omega \cdot \mathrm{m} .\) Its temperature coefficient of resistivity is \(0.0045\left(\mathrm{C}^{\circ}\right)^{-1},\) and this remains accurate even at high temperatures. The temperature of the filament increases linearly with current, from \(20^{\circ} \mathrm{C}\) when no current flows to \(2520^{\circ} \mathrm{C}\) at 1.00 A of current. (a) What is the resistance of the light bulb at \(20^{\circ} \mathrm{C} ?\) (b) What is the current through the light bulb when the potential difference across its terminals is \(120 \mathrm{~V} ?\) (Hint: First determine the temperature as a function of the current; then use this to determine the resistance as a function of the current. Substitute this result into the equation \(V=I R\) and solve for the current \(I .\) ) (c) What is the resistance when the potential is \(120 \mathrm{~V} ?\) (d) How much energy does the light bulb dissipate in 1 min when \(120 \mathrm{~V}\) is supplied across its terminals? (e) How much energy does the light bulb dissipate in 1 min when half that voltage is supplied?

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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