We assume your household supply is the common 117 volts.
Power = V2/R
60 = (117)2/R
60 R = (117)2
R = (117)2/60 = 228 ohms (rounded)
in a parallel circuit, the relationship of resistance is thus: 1/R1 + 1/R2 = 1/RT where R1 and R2 are two resistors in parallel and RT is the total resistance in the circuit: Using this, as more branches are added, the resistance decreases This is because there is more ways for the current to flow and thus using the analogy of a water flow with a constraint on it (resistor in electrical circuits), a whole load of streams with the same constraint on each will allow more water through than one large stream with the same small constraint on it.
I'm not quite sure what context you mean. V=IR to explain this in words you would simply say the voltage (V) of a circuit would be equal to the current (I) multiplied by the resistance(R). a practical example could be, if the resistance in a circuit was 2 and the current was 12 what is the voltage? well using Ohms law we would simply do (2 * 12) the voltage would be 24. [note, i do not know if that would work in real life it's just a example using numbers] you can aslo rearrange V=IR to work out how to calculate current and resistance V=IR I = V/R. Current equals votlage divided by resistance R=V/I. Resistance equals voltage divided by current hope this helps
On this calculation I am assuming that the light bulb is using a 120 volt source. Watts = Amps x Volts. Amps = Watts/Volts, 40/120 = .33 amps. R = Volts/Amps, 120/.33 = 363.6 ohms resistance in the 40 watt light bulb.
If it is a 1.5 volt bulb you can connect it through a series circuit directly to your battery. If the bulb is not made to run on 1.5 volts and needs another voltage, it may not light up!
Using a parallel circuit energy can be transferred through a parallel circuit.
"Volts" is electrical pressure applied to a circuit; whereas, "ohms" is electrical resistance to that pressure. One cannot determine ohms from voltage without knowing either the current (in "amps") or power (in "watts"). A normal 120V household circuit can handle a maximum of 20 amps, so using ohm's law of resistance = voltage / current, the minimum resistance required in a 120V household circuit would be 6 ohms. Any less than 6 ohms will cause the circuit breaker to trip.
Resistance of the circuit = (voltage across the circuit) divided by (current through the circuit)
You have to measure the voltage, and the current. The resistance is then calculated by using Ohm's Law.
Voltage drop is caused by circuit resistance
by using mirrors
A: A photocell is like a variable resistor it will change resistance with light the ratio of dark:light resistance decides the output. So yes it should work with 9volts
The net resistance can be found out using the algebraic sums f series and parallel connections. When there is no current flowing in the circuit the net resistance is infinite.
Firstly turn of the power before this test...Using a resistance or continuity tester you should get the following results:Short circuit: Very low resistance (nearly 0 ohms) or the bell will ring.Open circuit: Very high resistance (Somewhere in the range of Mega ohms) or the bell will not ring.The reason for this is because and open circuit has a gap in it (which has high resistance).The short circuit has wires that are crossed and so has a really low resistance.
It isn't. If you're using superposition, you open circuit current sources and short voltage sources; this is because the current source declares the current that will be flowing through that branch. Both current and voltage sources have a finite internal resistance.
Residence of one component varies overtime and from component to component.
As you add more bulbs to a series circuit that means that the bulbs are in series to one another, therefore the total resistance is the sum of the individual resistance of the bulbs. If you add bulbs of the same resistance,then the rate at which the resistance changes will increase in a constant manner provided the current source is not altered. For instance if the bulb you are using is rated 20v,60w, then the current passing via the bulbs in series is the square of the voltage divided by the power in this case the current is approximately 7amperes.
In a parallel circuit, each load added subtracts from total resistance. When one or more loads is removed from a parallel circuit, the total resistance is increased, reducing the total amperage draw. The less resistance a load has, the more current can pass through. This is part of Ohm's law. The mathematical equation that describes Ohm's law is: I=V/R , where I is the current in amperes, V is the potential difference in volts,and R is a circuit parameter called the resistance For example : The humble light-bulb is rated by the watts it uses. The amount of watts used by a light-bulb is calculated using Ohm's law. With the resistance of the bulb's filament and the voltage the bulb is designed to operate with, one can derive the amperage the bulb will draw. The amperage is then multiplied by the voltage to show wattage. Using Ohm's law : With the resistance of a 40watt 120volt light-bulb, only 0.33amps is able to pass through the bulb's 363ohm filament at 120volts. A lamp that has two 40watt bulbs inplace, and the two bulbs are in parallel, the circuit will have a resistance of 179ohms and draw 0.67amps which is 80watts at 120volts.