Yes, but then it would be a 'series-parallel' circuit, not a 'parallel' circuit!
Any circuit that even has more than one branch is a parallel one.
In a parallel circuit, the same (supply) voltage will appear across each branch. So, in your example, 12 V will appear across each of the 24-ohm resistors. To find the current through each resistor, then, you simply divide the supply voltage by the value of that resistor. Since the supply current is the sum of the two branch currents, to find the supply current, you simply add together the currents passing through each resistor.
A: There is no voltage drop running through in a parallel circuit but rather the voltage drop across each branch of a parallel circuit is the same
The rest of the lights in the system will remain illuminated. Except in that branch of the circuit. The parallel branch(s) get more current if the voltage potential remains the same.
If a 'parallel' circuit has more than one load in its (not "it's"!) branches, then it is not a parallel circuit, but a series-parallel circuit! To resolve the circuit, you must first resolve the total resistance of the loads within each branch.
No. What you are describing is 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) .
Voltage
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.
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.)