Physics - Electrical circuit?
820466
The voltage across each resistor will be 12 volts as each is effectively directly connected to the battery.\nIt is the current flow through the resistors which will be different.\nVolts/Ohms =Amps - that is\n12/7 amps or 12/5 amps now get your calculator out!\nRoyS
728562
Here is a good article. To understand them, you need to get a multimeter, and electronics projects book, a breadboard (to mount circuits on), and a selection of electronic parts from Radio Shack. Use batteries initially, and then when you are more confident a power supply and rectifier to do experiments, and measure voltage, current, and resistance values around the circuit.\nhttp://en.wikipedia.org/wiki/Electronics\nHyperphysics. (Google)\nhttp://www.mathworld.wolfram.com\nhttp://www.refdesk.com\n"You can do it electronics," Needham Heights, MA\nEnjoy!
731049
resistance is said to be one ohm if 1 ampere of current flows through the circuit across potential difference of 1volt  becuse resistance =voltage /current\na 1 ohm resistor gives obstruction to the current such that only 1A\ncurrent pass through it when potential difference is1V
783633
YOU ARE ASKING A QUESTION THAT IS ANSWERED BY SOLVING THE EQUATION (E=IR) OR OHMS LAW. IF YOU GIVE ME TWO OF THE  UNKNOWNS THE EQUATION CAN BE SOLVED FOR A NUMERIC ANSWER.   LETS ASSUME THE VOLTAGE(E) IS 100VOLTS.  THEN THE CURRENT(IR1) WOULD BE (100VOLTS/110 OHMS) E1/R1 OR 0.909 AMPS.\nTHE SITE BELOW GIVE A LENGTHY DESCRIPTION OF THE SIMPLE CONCEPT AND JUST SEARCHING "OHMS LAW" WILL YIELD DOZENS MORE. YOU WILL FIND THAT THE VOLTAGE APPLIED TO EACH RESISTOR IS THE SAME AND THE AMPERES THROUGH EACH RESISTOR IS INVERSELY PROPORTIONAL TO THE RESISTANCE.
858072
Resistance is the property in an electrical circuit.\n\nAny wire will have some resistance to flow of charges.\n\nThe property that gives hindrance to flow of charges is called resistance.\n\nIf there is any obstacle to the flow of charges then the one which gives this obstacle is said to possess some resistance.\n\nThe resistance is related by this simple equation E = I R.\n\nOr R = E/I. \n\nE is the potential difference and I is the rate of flow of charges.\n\nThe resistance is zero only when I is infinity.\n\nThe resistance is less when I is great for small values of E.\n\nThe resistance is large when I is small for a value of E.\n\nThe potential difference between the ends of a battery is full to its maximum when it is not connected to any external circuit.\n\nThat is when the current is zero the potential difference across the battery is maximum. This maximum potential difference is called E.M.F of the battery\n\nConsider a battery of 2 Volt.\n\nWhen this is connected to a resistor of 2 Ohm the current we expect is 2/2 = 1 ampere. \n\nBut in practice this will not happen.\n\nThe current will be less than 1A. Suppose that the current is 0. 95 A.\n\nAt the same time if we measure the potential difference it will NOT be 2 volt but it will be only 1.9 volt. \n\nThus there is a loss of 0.1 volt.\n\nThe chemicals inside the battery produce this 0.1 volt opposing the 2 volt ( when there is no flow of charges inside the battery).\n\n\n \n\nReturning to the current consideration, we used a battery of 2 volt E.MF and a resistor of resistance 2 ohm. But the current is 0.95 A instead of 1 A.\n\nIf the total resistance were 2.105 ohm instead of 2 ohm , then the current in the circuit will be 2 / 2.105 = 0.95 A.\n\nThus we see that the battery is having some INTERNAL resistance of 0.105 ohm.\n\nThis resistance is due to the production of opposite emf of 0.1 volt to the actual emf of 2 volt (2- 01 = 1.9 V). The potential difference across the battery has reduced to 1.9V from 2 V.\n\nThus the resistance is only a property and there not be any resistor inside the battery.\n\nThis internal resistance is not a constant one and will vary according to the external resistances used in the circuit and other factors.
702210
If you know mechanical terms, think of it like this. Your electrical circuit has become a closed fluid system (city water supply) instead.\n\nCurrent is the "mass flow rate" of electrons.\n\nVoltage is the "system pressure".\n\nYou can have a pressurized water pipe with no fluid movement (Your faucet). In an equivelant electrical circuit, you would have a Voltage (Potential) with no Current inside the "Circuit".\n\nYou can have fluid/ electrons moving through a pipe/ wire but inevitably, something has to be helping it through (a Voltage difference or a Pressure difference).\n\nTherefore:\nYou can have Voltage/Potential (a pressure) without Current (flow). But you can't have Current (flow) without some voltage (pressure difference). \n\nWhen the breaker/ switch opens, current ceases to exist. The Voltage (charge) is not reliant on the existence of a current.\n\nWhen the flow of electrons stops, current ceases to exist. Regardless of the existing Potential you can still find, the nonexistant current cannot be defined or measured.
438682
sure. go to a science store or look up schematics on the web.
655128
Assuming that we have 2 series resistors R1 and R2. And the voltage applied on them is V.\n\nTo get voltage applied on R1 (V1)= V * R1/(R1+R2)\nand so V2= V * R2/(R1+R2)
238179
Resistance=Voltage/Amperes
31337
It is DC voltage.\n\nhttp://www.ecs.umass.edu/ece/allerton/papers1999/36/
768635
Well we are given voltage, and power.  We are asked to find the current. Well the formula we have to use here is P=VI.\nRearranging this formula we get; I=P/V\n\nHowever; there is only one thing wrong with this problem.\nWe do not know what the voltage is correctly.\nIf this diode is a ideal diode. An ideal diode acts like a short circuit with no voltage drop across it. Then we would use .8V as the Voltage Source across the diode.\nIdeal Diode:\nI = .05 W/.8V\nI = 62.5mA\n\n\nIf it is not an ideal diode and there is a .75V drop across the diode then we would use .05V as the Voltage Source across the diode.\nI = .05W / (.8V - .75Vdiode drop)\nI = .05W / (.05V)\nI = 1A\n\nHere this might give your more information!\nhttp://en.wikipedia.org/wiki/Diode_modelling
964775
electropneumatic circuit \n\nhttp://www.festo-didactic.com/ov3/media/customers/1100/0000607001156324537.pdf\n\n:>peace
15184
Using Ohm's Law, V=IR, we have 10=I(5).  Solving for I, the current is 2 Amps.
471894
a drawbridge or a circuit??
813641
using KCL (current law), A is corrent.\nusing KVL (voltage law), B is correct\nC doesn't make any sense, and you can support that with the definition of the KCL.
788402
You can find the current flowing through a Wheatstone bridge by using Kirchoff's Loop Rules and Junction Rules.      \n\nThe sum of the Voltage Sources - Voltage Drops = 0 in a closed circuit.  \n\nThis setup can be solved using a matrix or a CAS. \n\nGood Luck
782944
I assume you mean a parallel circuit?\n\nIn electronics, this is a circuit in which there are two different paths for the electrical current to flow through at the same time.\n\nImagine a battery with two wires coming out from the positive side and two wires coming out from the negative side.  There is one resistor connected between the first positive wire and the first negative wire, and a second resistor connected between the second positive wire and the second negative wire.  Some current from the battery flows through the first resistor, and some current from the battery flows through the second resistor.\n\nBasic rules for parallel circuits are as follows:\nVoltages across each "leg" (in the example, each resistor) of the circuit are always equal to one another.\nCurrent through the battery is equal to the sum of the currents in each of the legs.\nPower is equal to the sum of the powers in each of the legs.