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Before bendable tungsten filaments were developed, Thomas Edison used carbon filaments in his light bulbs. Though carbon has a very high melting temperature \(\left(3599^{\circ} \mathrm{C}\right)\) its sublimation rate is high at high temperatures. So carbonfilament bulbs were kept at lower temperatures, thereby rendering them dimmer than later tungsten-based bulbs. A typical carbon-filament bulb requires an average power of \(40 \mathrm{~W}\), when 110 volts is applied across it, and has a filament temperature of \(1800^{\circ} \mathrm{C}\). Carbon, unlike copper, has a negative temperature coefficient of resistivity: \(\alpha=-0.0005^{\circ} \mathrm{C}^{-1}\) Calculate the resistance at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) of this carbon filament.

Short Answer

Expert verified
Answer: The resistance of the carbon filament at room temperature is approximately \(571.73 \ \Omega\).

Step by step solution

01

Calculate resistance at the operating temperature

Using the formula for power: \(\text{Power} = \frac{\text{Voltage}^2}{\text{Resistance}}\), we can rearrange to solve for resistance: \(\text{Resistance} = \frac{\text{Voltage}^2}{\text{Power}}\) Now, we can plug in the given power of 40 W and voltage of 110 V: \(\text{Resistance}_{1800^{\circ} \mathrm{C}} = \frac{(110 \mathrm{~V})^2}{40 \mathrm{~W}} = \frac{12100 \mathrm{~V}^2}{40 \mathrm{~W}} = 302.5 \ \Omega\)
02

Determine the change in temperature

We need to find the change in temperature between the operating temperature and room temperature: \(\Delta T = T_{\text{operating}} - T_{\text{room}} = 1800^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 1780^{\circ} \mathrm{C}\)
03

Calculate the resistance at room temperature

Using the formula for temperature dependence of resistance: \(R_{20^{\circ} \mathrm{C}} = R_{1800^{\circ} \mathrm{C}} \left[1 - \alpha \cdot \Delta T\right]\) Plugging in the values found in Steps 1 and 2, and given \(\alpha = -0.0005^{\circ} \mathrm{C}^{-1}\): \(R_{20^{\circ} \mathrm{C}} = 302.5 \ \Omega \left[1 - (-0.0005^{\circ} \mathrm{C}^{-1}) \cdot 1780^{\circ} \mathrm{C}\right]\) \(R_{20^{\circ} \mathrm{C}} = 302.5 \ \Omega \left[1 + 0.89\right]\) \(R_{20^{\circ} \mathrm{C}} = 302.5 \ \Omega \cdot 1.89 = 571.725 \ \Omega\) Therefore, the resistance of the carbon filament at room temperature is approximately \(571.73 \ \Omega\).

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Key Concepts

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

Electrical Resistance
Electricity is all about the movement of electrons, and resistance is what opposes this movement. Think of it like trying to slide down a slide covered in sticky syrup; the syrup is like resistance—it's what makes it hard to slide down smoothly. In technical terms, electrical resistance measures the degree to which a material hinders the flow of electric current.

Materials with higher resistance make it harder for the current to flow, which is why we don't use something like rubber to make electrical wires; it has too much resistance, and the electrons can't flow through easily. Metals like copper or aluminum, on the other hand, have a much lower resistance and are great for wires. The resistance of a material depends on its composition, length, thickness, and even the temperature, which can cause the resistance to change.
Power Calculation
When you're about to charge your phone, you plug it into the wall without thinking much about it, right? But there's some cool science happening there!

Power calculation is like working out how much energy your phone is getting from that outlet every second. The power (measured in watts) tells us the rate at which energy is used or delivered. We can calculate it by taking the square of the voltage (yes, voltage is a big deal—it's like the pushing force for electrons) and then dividing it by the resistance (

For example, a light bulb that requires 40 watts to light up properly needs a good balance of voltage and resistance to work right. If the voltage is too low or the resistance too high, the light bulb won't shine as brightly, just like how your phone won't charge well with a weak charger.
Temperature Dependence of Resistance
Just like how your fingers get all clumsy when they're cold, materials can act differently when the temperature changes. Specifically, the resistance of materials changes with temperature—a concept crucial in designing all sorts of electronic gadgets.

In most metals, as the temperature goes up, the atoms jiggle around more, which makes it tougher for the electrons to slip through the material, resulting in higher resistance. This is why your phone might get a bit wonky if it overheats—the internal resistance changes might muddle its performance.

However, some materials, like our friend carbon used in the old light bulbs, exhibit negative temperature coefficients of resistivity. That means their resistance actually goes down as the temperature goes up. It's like resistance is moving backward: the warmer the carbon gets, the easier it is for electricity to flow through it. Strange, but true - science is full of surprises!

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Most popular questions from this chapter

Two identical light bulbs are connected to a battery. Will the light bulbs be brighter if they are connected in series or in parallel?

Which of the following wires has the largest current flowing through it? a) a 1 -m-long copper wire of diameter \(1 \mathrm{~mm}\) connected to a \(10-V\) battery b) a \(0.5-\mathrm{m}\) -long copper wire of diameter \(0.5 \mathrm{~mm}\) connected to a 5 -V battery c) a 2 -m-long copper wire of diameter \(2 \mathrm{~mm}\) connected to a \(20-V\) battery d) a \(1-\mathrm{m}\) -long copper wire of diameter \(0.5 \mathrm{~mm}\) connected to a 5 -V battery e) All of the wires have the same current flowing through them.

A copper wire has a diameter \(d_{\mathrm{Cu}}=0.0500 \mathrm{~cm}\) is \(3.00 \mathrm{~m}\) long, and has a density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). As shown in the figure, the copper wire is attached to an equal length of aluminum wire with a diameter \(d_{\mathrm{A} \mathrm{I}}=0.0100 \mathrm{~cm}\) and density of charge carriers of \(6.02 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). A current of 0.400 A flows through the copper wire. a) What is the ratio of the current densities in the two wires, \(J_{\mathrm{Cu}} / J_{\mathrm{Al}} ?\) b) What is the ratio of the drift velocities in the two wires, \(v_{\mathrm{d}-\mathrm{Cu}} / v_{\mathrm{d}-\mathrm{Al}} ?\)

A resistor of unknown resistance and a \(35-\Omega\) resistor are connected across a \(120-\mathrm{V}\) emf device in such a way that an 11 -A current flows. What is the value of the unknown resistance?

What would happen to the drift velocity of electrons in a wire if the resistance due to collisions between the electrons and the atoms in the crystal lattice of the metal disappeared?

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