No. What you are describing is a series-parallel circuit, not a parallel circuit.
The current in the circuit will depend on how the three resistors are wired. Series? Parallel? Series parallel? With the resistors in series, 3, 2 and 4 ohms will add to 9 ohms. As I = E/R, I = 9 V / 9 ohms = 1 A. With the resistors in parallel, the 3, 2 and 4 ohm resistors will draw 3 A, 4.5 A and 2.25 A respectively, and the total current will be the sum of the branch currents, or 3 A + 4.5 A + 2.25 A = 9.75 A. There are 3 different series parallel circuits possible, and more investigation will be necessary to solve for them.
Any circuit that even has more than one branch is a parallel one.
In a parallel circuit, the current flow is independent in each branch.
Increases
Two receptacles on a branch circuit, each in parallel, both in series with the circuit breaker. The blower motor, ignition transformer, and oil solenoid on an oil burner, each in parallel, all in series with the acquastat (water temperature control switch).
Yes, but then it would be a 'series-parallel' circuit, not a 'parallel' circuit!
2 amps
The ratio of current flow through individual branches of a parallel circuit is inversely proportional to the ratio of resistance of each branch.
Any circuit that even has more than one branch is a parallel one.
The current in the circuit will depend on how the three resistors are wired. Series? Parallel? Series parallel? With the resistors in series, 3, 2 and 4 ohms will add to 9 ohms. As I = E/R, I = 9 V / 9 ohms = 1 A. With the resistors in parallel, the 3, 2 and 4 ohm resistors will draw 3 A, 4.5 A and 2.25 A respectively, and the total current will be the sum of the branch currents, or 3 A + 4.5 A + 2.25 A = 9.75 A. There are 3 different series parallel circuits possible, and more investigation will be necessary to solve for them.
Yes, the total power dissipated through the circuit is equal to the sum of the power of each branch in a parallel circuit.
It's usually referred to as one leg of the circuit.
The least amount of current will flow through the branch of a parallel circuit that has the most resistance.
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) .
According to Kirchhoff's Current Law, the sum of the individual branch currents must be equal to the total current before (and after) it branches.
Voltage
Yes. The voltage across every branch of a parallel circuit is the same. (It may not be the supply voltage, if there's another component between the power supply and either or both ends of the parallel circuit.)