The battery has 6 volts across its terminals. The way to discover it is to apply Ohm's law. It (Ohm's law) comes in 3 "flavors" that look a bit different but all say exactly the same thing. Here they are: E = I x R [Voltage equals current times resistance.] I = E/R [Current equals voltage divided by resistance.] R = E/I [Resistance equals voltage divided by current.] In these equations, voltage is E, current is I and resistance is R. They are measured in units of volts, amperes (or amps) and ohms, respectively. In your problem, we have the resistance (R) and the current (I). We need to find the voltage (E), and the formula E = I x R is the logical choice to discover the voltage. As E = I x R here, E = 0.75 x 8 = 3/4 x 8 = 6 volts. Piece of cake.
Voltage = Current x Resistance
E = I x R
E = 8amps x 45 ohms = 360 volts
V = I.R
V = 1.5 x 8
V = 12V
Use the equation V = I * R 20 = I * 5000 I = 20 / 5000 I = 0.004 Amps The answer to your question is 4 mA of current will flow through the resistor.
0.81 APEX
A very very tiny amount of the current that would normally flow through the resistor instead flows through the voltmeter, allowing it to make its measurement. For most purposes this very very tiny amount of current can be completely ignored.
The current in each resistor in a series circuit is the same. Kirchoff's Current Law states that the sum of the currents entering a node must add up to zero. The connection between two resistors in a series circuit is a node. The current entering the node from one resistor is equal to the current leaving the node into the next resistor.
3 ohms. 9 volts across a 3 ohm resistor becomes 9/3 or 3 amps.
All the way along when the crocodile clip is connected to a resistor, when the other end of the resistor is connected to the other side of the battery.
12 milliamps
Voltage = Current * ResistanceVoltage = 12VResistance = 10 ohmsCurrent = Voltage/ResistanceCurrent = 12V/10 ohmsCurrent = 1.2 Amps
yes... this is possible if a diode i connected in reverse bias with a battery and a resistor for example. A diode in reverse bias means its anode will be connected to positive terminal of the battery and its cathode to the negative terminal of the battery. In such a case, minimal current flows through the circuit which can be neglected.
A circuit with a 2 ohm resistor and a 4 ohm resistor in series with a 12 volt battery will have 2 amps flowing through each resistor. The current is the same in each resistor because they are in series, and a series circuit has constant current throughout.
True
as the given cells have the same current flowing in through them (current flowing through the cells connected in series is equal to the current flowing when connected in parallel ) equate the formula's of cells connected in series and cells connected in parallel.thus by equating we get the value of the internal resistor as 2 ohms.
1amp
6
Current will always flow in both resistors, but the one with the lower resistance will have more current flow through it. The value of the current in each resistor is calculated by dividing the voltage of the source by the resistance of the individual resistor. As long as the capability of the power source isn't exceeded, the current through each resistor isn't affected by the presence of the other resistor. Said another way, if two resistors are connected in parallel across a source, neither one "cares" that the other resistor is connected across the source. The two resistors work independently.
2
The resistors are 5, 10, and 15 ohms.The current through the 5-ohm resistor is E/R = 15/5 = 3 Amp.The current through the 10-ohm resistor is E/R = 15/10 = 1.5 Amp.The current through the 15-ohm resistor is E/R = 15/15 = 1 Amp.Their total effective resistance in parallel is the reciprocal of [ (1/5) + (1/10) + (1/15) ] =the reciprocal of [ (6/30) + (3/30) + (2/30) ] = the reciprocal of [ (11/30) ] = 30/11 ohms .The total current drawn from the battery is E/R = (15)/(30/11) = (15 x 11/30) = 11/2 = 5.5 Amp.Note:The 5-ohm resistor is dissipating 45 watts.The 10-ohm resistor is dissipating 22.5 watts.The 15-ohm resistor is dissipating 15 watts.The poor battery is delivering 82.5 watts.None of this is going to last very long at all.Most likely, the battery has already expired,and/or the 5-ohm resistor has already exploded,while we've been here playing with our calculators.