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If all the bulbs are connected in parallel, and there is enough current, yes, the brightness will be the same. The voltage (which is the amount of energy in every charge), remains the same for all bulbs
Kirchoff's Current Law: The signed sum of the currents entering a node is zero. Assume the top of the battery is a node. The current entering it (from the battery) is equal to the sum of the currents leaving it (to the branches). This adds up to zero.
They will both last for the same amount of time. In both situations all of the battery's power is being used, but in parallel, you can separately control the individual loads that you put on the circuit with switches. Hope that helps
They provide a positive shut off and offer the least amount of resistance to flow when fully open.
An ammeter.
The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).
The other bulbs continue to glow, with the same power as before. The total current and power decreases by the amount lost to the one bulb.
(mA of current the circuit draws from the battery when it's running) multiplied by (number of hours you want the circuit to operate from the battery) is equal to the absolute minimum mAh rating the battery must have.
Kirchoff's Current Law: The signed sum of the currents entering a node is equal to zero. This means that the current leaving the battery is equal to the sum of the currents in the branches of the circuit.
If at battery,parallel circuit shorts then equivalent resistance of circuit becomes approximately 0 Ohms,and therefore as current follows low resistance path infinite amount of current due to low resistance will flow through the wire so,entire parallel circuit will short out,but wire will burn and battery may get damaged. Name:Sumit Karnik.
If all the bulbs are connected in parallel, and there is enough current, yes, the brightness will be the same. The voltage (which is the amount of energy in every charge), remains the same for all bulbs
Kirchoff's Current Law: The signed sum of the currents entering a node is zero. Assume the top of the battery is a node. The current entering it (from the battery) is equal to the sum of the currents leaving it (to the branches). This adds up to zero.
a single circuit has only one source of energy, like a battery compartment (no matter the amount of batteries) a double circuit has two sources of energy, like two batter compartments
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.
There will be plugs on the top of the battery that will unscrew. Most new batteries are maintenence free so adding water is not needed if you feel your not having the right amount of charge I suggest you have the battery tested before you cause harm to yourself
When adding and subtracting a constant amount means that that amount will increase. The amount will increase dew to adding each number.
If a battery sends a current of 10A through a circuit for one hour how many coulombs will flow through the circuit?