2 resistors 3 Ω and 6 Ω are connected in parallel across a 6 V battery. What is the current flow through the 3 Ω resistor?
6 A
3 A
4 A
2 A
Both B and D sound possible to me..
6 = i (3)
i = 2
I1=U/R1=6.3 =2 A
To determine the current flow through the 3 Ω resistor, we need to apply Ohm's law and the rules for resistors in parallel.
In a parallel circuit, the voltage across each resistor is the same. Therefore, the voltage across the 3 Ω resistor is 6 V, as it is connected across a 6 V battery.
To find the current flowing through a resistor, we can use Ohm's law: I = V/R, where I is the current, V is the voltage, and R is the resistance.
Using Ohm's law, we can calculate the current flowing through the 3 Ω resistor:
I = V/R = 6 V / 3 Ω = 2 A
Therefore, the current flow through the 3 Ω resistor is 2 A. So, the correct answer is D) 2 A.
To determine the current flow through the 3 Ω resistor, we need to apply Ohm's Law and use the concept of parallel resistors.
When resistors are connected in parallel, the voltage across each resistor is the same, while the total current is equal to the sum of the currents through each individual resistor.
First, we need to calculate the equivalent resistance of the parallel combination of the 3 Ω and 6 Ω resistors. The formula for calculating the equivalent resistance of two resistors in parallel is:
1/Requivalent = 1/R1 + 1/R2
Plugging in the values, we get:
1/Requivalent = 1/3 Ω + 1/6 Ω
1/Requivalent = (2 + 1)/6 Ω
1/Requivalent = 3/6 Ω
1/Requivalent = 1/2 Ω
Re-arranging the equation:
Requivalent = 2 Ω
Now that we have the equivalent resistance, we can use Ohm's Law to calculate the current flowing through the circuit:
I = V / R
where I is the current, V is the voltage, and R is the resistance.
In this case, the voltage across the resistors is 6 V (as stated in the question), and the resistance is 3 Ω. Plugging in these values, we get:
I = 6 V / 2 Ω
I = 3 A
Therefore, the current flow through the 3 Ω resistor is 3 A. Hence, option B is correct.