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This problem involves two steps. First we need to figure out the amount of current in each light bulb. Then we use that value to find our two different resistances using Ohm's Law, and combine them to find our total resistance of the two bulbs.

For the first step, we need to know that the 30 and 40 watts they give us describe the light bulb's power, which is the rate at which they transform electricity into light and heat. There is an equation in physics which states that the power of a light bulb or other resistor (any object that takes up electricity in a circuit), is equal to its current, measured in amps, multiplied by the volts supplied by the battery. In equation form, P=IV, where P=power, I=current, and V=volts. We need I to find our resistance for the bulbs later.

The power of the first bulb is 30 watts, and the volts equals 120 V, so 30=I(120). I works out to be .25 amps for the first bulb. The second bulb has a power of 40 watts, but receives the same amount of volts as the first bulb because they are in the same circuit. So for the second bulb, 40=I(120), and I works out to be .333 amps.

Now because we have the resistances, we can use Ohm's Law for both bulbs. Ohm's Law states that the current equals the the voltage divided by the resistance, or I=(V/R). To make this easier, I'll change the equation a bit. We can multiply both sides by R to get IR=V, and then divide by I to get the new equation, R=V/I. This makes it easier for us to find our resistance directly with our values.

Then we simply plug in our I values which we found and the voltage of the battery to figure out the resistances. The first bulb's equation is R=120/.25, and the second bulb's equation is R=120/.333. Solving for the R's, the first bulb has a resistance of 360 Ohms, while the second bulb has a resistance of 480 Ohms.

If you just wanted the individual resistances, then those are your answers. But because you told me they were in series, it seems to me that you want their total resistance. We can find the total resistance for resistors in series by simply adding the resistances up. In this case, our total resistance would be 360 Ohms plus 480 Ohms, or 840 Ohms.

Take a look at the links below for more information. The first one explains the relationships between current, voltage, and resistance, and the second link discussed finding resistance in series and in parallel as well.

Q: What is the resistance of two light bulbs in a series if one is 30watts and the other is 40watts with a 120volt battery?

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A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)The amount of lost volts depends on the current being drawn:The less resistance a circuit has, the more current is drawn, because it's easier to flow.Example:If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.If the circuit has high resistance, it draws a small current and there are fewer lost volts.This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.

The energy delivered by a battery would depend on-- the battery's voltage-- the resistance of the load connected across its output terminals-- the length of the time the load is connectedThe power delivered by the battery is [ (voltage)2 divided by (load resistance) ].The total energy delivered by the battery is [ (power) multiplied by (time the load is connected) ].

Use Ohm's Law. Solving for current:I = V/R (current = voltage / resistance)

A resistance of 3 ohms connected between the terminals of a 9-volt battery will result in a current of 3 Amperes. If the battery is one of those little ones with snaps on top, it may be able to produce 3 amperes of current for about 3 seconds before it rolls over and totally dies.

In a complex circuit with various elements (resistors, capacitors etc.) and one battery, the various circut elements contribute to draw a certain amount of current "I"from the battery at some terminal voltage "V". The "equivalent" resistance of the various circuit elements is that resistance "R" which will draw the same current , at the same terminal voltage, as the complex circuit. So to find "R" you simply imagine replacing the complex circuit with "R" by attaching "R" across the terminals of the battery and use Ohms law to find "R" , demanding "I" and "V" are the same. So then R = V/I.

Related questions

No, it is desirable for a battery to have a low internal resistance.

The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).

A 24Whr battery lasts 24hr if the system is using 1 Watts/hr. So it depends on how many watts the system uses. Typical netbooks may use about 30watts/hr so you may get about an hr on a full charge....

The value of internal resistance of 1.5 volt battery is 0.5 ohms.

There is internal resistance in a battery because a battery is not an ideal voltage source. It may be close, but it is not ideal. As a result, analytically, there will be some series resistance, resistance which places a limit on the maximum current that the battery can provide. While no battery is ideal, most are sufficiently ideal to not require any consideration of the internal resistance. If your circuit is dependent on the internal resistance of a battery, then it is probably not well suited for that application.

Yes

That will depend on the internal resistance of the battery. I = E / R Where I is the current, E is the open circuit battery voltage, and R is the internal resistance of the battery.

You have to imagine the internal resistance as being in parallel with any load you connect. You get the maximum possible current when the load is zero. In this case, just apply Ohm's Law. That is, divide the voltage by the internal resistance.

011.1

The apmeres depend on the resistance of the circuit. The battery will be a certain voltage, and dividing the voltage by the resistance gives you amperes. V = I*R

Internal resistance is approximately equal to 94.667

A battery is rated to supply a certain number of volts. However, it actually supplies less, because they are "lost" as the current has to get out of the battery in the first place.(The battery has internal resistance)The amount of lost volts depends on the current being drawn:The less resistance a circuit has, the more current is drawn, because it's easier to flow.Example:If the circuit has little resistance, it draws a large current and the battery's internal resistance causes more lost volts.If the circuit has high resistance, it draws a small current and there are fewer lost volts.This is why when you short-circuit a battery (give it hardly any resistance to go through) it heats up and may explode. A large current is drawn and all the volts are used by the battery's internal resistance.