Use Ohm's Law. Solving for current:I = V/R
(current = voltage / resistance)
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.
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.
"http://www.tpub.com/neets/book1/chapter3/1-26.htm" This site explains how to calculate the resistance, but it decreases the resistance when you add more.
Current = voltage/resistance If those are the only components in the circuit, then Current = 9/12 = 0.75 Ampere = 750 mA
240 ohms
The voltage of the battery, and the resistance of the circuit (including the resistance of the wire and the internal resistance of the battery).
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.
100.0 (apex)
To solve any D.C. circuit by using Thevenin Theorem,First of all load resistance RL is disconnected from the circuit and open circuit voltage across the circuit is calculated (known as Thevenin equivalent voltage)Secondly, the battery is removed by leaving behind its internal resistance. Now we calculate equivqlent resistance of the circuit ( called Thevenin equivalent resistance).Now we connect Thevenin Voltage in series with Equivalent resistance of the circuit and now connect load resistance across this circuit to calculate current flowing through the load resistance.Whereas in the case of using Norton theorem, we again remove the load resistance if any, and then short circuit these open terminals and calculate short circuit current Isc.Second step is same as in Thevenin theorem i.e. remove all sources of emf by replacing their internal resistances and calculate equivqalent resistance of the circuit.Lastly, join short circuit current source in parallel with equivalent resistance of the circuit. Now, we can calculate votage across the resistance which was connected in parallel with Isc.So, by knowing the open circuit voltage, we can calculate current flowing the resistance and on the other hand , by knowing the short curcuit current , we can calculate voltage across the resistance.
The resistance of a series circuit is simply the sum of the individual resistors.
Hi, there. A battery is a power supply, a source of potential difference which drives current. In itself, a battery is not a circuit, but if you attach it to a load (a resistance), then a current will form and a circuit is made!
It does not contain unidirectional outputAnswerA purely resistive circuit is an 'ideal' circuit that contains resistance, but not inductance or capacitance.
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.
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.
Batteries are rated as ampere/hour any circuit that draws power from it effects it. The lower the internal resistance of the circuit the shorter the useful battery life as discharged.
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.
It does not contain unidirectional outputAnswerA purely resistive circuit is an 'ideal' circuit that contains resistance, but not inductance or capacitance.