Write the power rating of any one appliance in your house and then calculate: i. Resistance of the appliance. ii. Maximum current limit of the appliance. iii. Electrical energy consumption of the appliance in kWh when operated for 1hour. iv. The cost of electrical energy used if the cost of electrical energy is Rs. 6 per kWh.
Answers
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Explanation:
Power is associated by many people with electricity. Knowing that power is the rate of energy use or energy conversion, what is the expression for electric power? Power transmission lines might come to mind. We also think of lightbulbs in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb. (See Figure 1(a).) Since both operate on the same voltage, the 60-W bulb must draw more current to have a greater power rating. Thus the 60-W bulb’s resistance must be lower than that of a 25-W bulb. If we increase voltage, we also increase power. For example, when a 25-W bulb that is designed to operate on 120 V is connected to 240 V, it briefly glows very brightly and then burns out. Precisely how are voltage, current, and resistance related to electric power?
Which of the lightbulbs, the 25-W bulb (upper left) or the 60-W bulb (upper right), has the higher resistance? Which draws more current? Which uses the most energy? Can you tell from the color that the 25-W filament is cooler? Is the brighter bulb a different color and if so why? (credits: Dickbauch, Wikimedia Commons; Greg Westfall, Flickr) (b) This compact fluorescent light (CFL) puts out the same intensity of light as the 60-W bulb, but at 1/4 to 1/10 the input power. (credit: dbgg1979, Flickr)
Electric energy depends on both the voltage involved and the charge moved. This is expressed most simply as PE = qV, where q is the charge moved and V is the voltage (or more precisely, the potential difference the charge moves through). Power is the rate at which energy is moved, and so electric power is
Recognizing that current is I = q/t (note that Δt = t here), the expression for power becomes
P = IV
Electric power (P) is simply the product of current times voltage. Power has familiar units of watts. Since the SI unit for potential energy (PE) is the joule, power has units of joules per second, or watts. Thus, 1 A ⋅V= 1 W. For example, cars often have one or more auxiliary power outlets with which you can charge a cell phone or other electronic devices. These outlets may be rated at 20 A, so that the circuit can deliver a maximum power P = IV = (20 A)(12 V) = 240 W. In some applications, electric power may be expressed as volt-amperes or even kilovolt-amperes (1 kA ⋅V = 1 kW). To see the relationship of power to resistance, we combine Ohm’s law with P = IV. Substituting I = V/R gives P = (V/R)V=V2/R. Similarly, substituting V = IR gives P = I(IR) = I2R. Three expressions for electric power are listed together here for convenience:
Note that the first equation is always valid, whereas the other two can be used only for resistors. In a simple circuit, with one voltage source and a single resistor, the power supplied by the voltage source and that dissipated by the resistor are identical. (In more complicated circuits, P can be the power dissipated by a single device and not the total power in the circuit.) Different insights can be gained from the three different expressions for electric power. For example, P = V2/R implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in P = V2/R, the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.
EXAMPLE 1. CALCULATING POWER DISSIPATION AND CURRENT: HOT AND COLD POWER
(a) Consider the examples given in Ohm’s Law: Resistance and Simple Circuits and Resistance and Resistivity. Then find the power dissipated by the car headlight in these examples, both when it is hot and when it is cold. (b) What current does it draw when cold?
Strategy for (a)
For the hot headlight, we know voltage and current, so we can use P = IV to find the power. For the cold headlight, we know the voltage and resistance, so we can use P = V2/R to find the power.
Solution for (a)
Entering the known values of current and voltage for the hot headlight, we obtain
P = IV = (2.50 A)(12.0 V) = 30.0 W.
The cold resistance was 0.350 Ω, and so the power it uses when first switched on is
Discussion for (a)
The 30 W dissipated by the hot headlight is typical. But the 411 W when cold is surprisingly higher. The initial power quickly decreases as the bulb’s temperature increases and its resistance increases.
Hope it helps...
Pls mark as brainliest.
Pls mark as brainliest.
Pls mark as brainliest.
Pls mark as brainliest.
