so i have a circuit (series) where the the emf=12 V and R1=2 ohms and R2=0 ohms.
Basically, theres's a square. The battery voltage(12V) is on the left side. The R1=2 is on the top side of the square and R2=0 ohms is on the right side of the square.
The total current of the series is 6 A and the voltage is 12 A.
I know in series, the current is the same for all resistors so since R2=0 ohms, will the voltage for R2 and Current for R2 be zero while voltage for R1 is 12 V and current for R1 is 6 Amps.
or will current for R2 be 6 Amps as well.
R2 will pass all the current (6A) with no voltage across it.
In a series circuit, the current remains the same throughout all the resistors. However, since R2 has a resistance of 0 ohms, it can be considered a short circuit. In a short circuit, the current takes the path of least resistance, effectively bypassing R2.
As a result, all of the current will flow through R1, which has a resistance of 2 ohms. Therefore, the current through R1 will be 6 Amps, and the current through R2 will be 0 Amps.
Now let's consider the voltage across each resistor. To find the voltage across a resistor in a series circuit, you can use Ohm's Law, which states that V = I * R, where V is the voltage, I is the current, and R is the resistance.
For R1, with a current of 6 Amps and a resistance of 2 ohms, the voltage across it can be calculated as V1 = I * R1 = 6 A * 2 Ω = 12 V.
For R2, since there is no current flowing through it, the voltage across it will be zero.
So, in summary, the current through R2 will be 0 Amps, the voltage across R2 will be 0 V, the current through R1 will be 6 Amps, and the voltage across R1 will be 12 V.