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.
If three lamps are connected in parallel and one blows out, the current in the other two does not change. This assumes that the net change in total current does not cause the power source to change voltage.
A current that runs parallel to the shoreline is called a rip tide or rip current.
total resistance is each resistor added together
The current splits up and takes as many paths as there are available. Every path that exists conducts current. There's no such thing as a parallel branch of a circuit that has no current through it when others do have current through them..
You add up the currents in each branch. The current in each branch is just (voltage acrossd the parallel circuit)/(resistance of that branch) . ==================================== If you'd rather do it the more elegant way, then . . . -- Write down the reciprocal of the resistance of each branch. -- Add up the reciprocals. -- Take the reciprocal of the sum. The number you have now is the 'effective' resistance of the parallel circuit ... the single resistance that it looks like electrically. -- The total current through the parallel circuit is (voltage acrossd the parallel circuit)/(effective resistace of the parallel circuit) .
If a 9.0 volt battery is connected to a 4.0-ohm and 5.0-ohm resistor connected in series, the current in the circuit is 1.0 amperes. If a 9.0 volt battery is connected to a 4.0-ohm and 5.0-ohm resistor connected in parallel, the current in the circuit is 0.5 amperes.
When a resistor is added the current goes down, that is expressed in the equation current= voltage/ resistance
The 12V battery connected to the 2.4 Ohm combination will supply 12/2.4 or 5A. The individual currents will be 12/3 or 4A for the 3 Ohm resistor, and 12/12 or 1A for the 12 Ohm resistor. The 2.4 Ohm parallel combination is obtained from a simple product-over-the-sum calculation.
half of the current flowing thru resistor 1.... V=IR.
the voltage across that resistor will increase if it is in series with the other resistors. the current through that resistor will increase if it is in parallel with the other resistors.
The voltage of a battery goes as the current times the resistance (V=IR). Because the voltage is being held constant, the resistor that draws the most current will have the lower resistance.
When a capacitor and resistor are connected in parallel in a circuit, the behavior changes in that the capacitor stores and releases electrical energy while the resistor controls the flow of current. This combination can affect the overall impedance and time constant of the circuit, leading to changes in the voltage and current characteristics.
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.
It depends on the purpose for installing the resistor. If the intent is to decrease current flow, the resistor must be connected in series with the load. If the purpose is to increase current flow, the resistor must be connected in parallel with the load. To connect a resistor in series, connect the resistor to one side of the power source, in line with the load. This will decrease circuit current flow. To connect a resistor in parallel, connect the resistor between the positive and negative sides of the power source, which will effectively connect the resistor across the load . This will increase current flow through the circuit. However, before connecting a component in parallel, make sure the increase in current flow will not exceed the current rating of the circuit or fuses/breakers will blow.
In simple way resistor bank contains number of resistors in series or parallel combination. They are connected in parallel to decrease the resistance and increase current rating and power dissipation.And they are connected in series to increase resistance and power dissipation.
1amp
12 milliamps