A circuit contains a 12V power source, and a 3ohms, 4ohms, and 6ohms, resistor. How should the resistors be arranged for the current through the battery to be 2A?
Please help!THANKS!
you want a total of six ohms. The easy way is to put the six ohm resistor in the circuit, and then dangle the other two resistors off a connection, with the other end unconnected.
The other way: What is the resistance of the six and three ohm in parallel?
To arrange the resistors in such a way that the current through the battery is 2A, you need to apply Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) across the circuit divided by the resistance (R) of the circuit.
First, you need to calculate the total resistance of the circuit. In this case, the resistors are arranged in series, so you can simply add their resistances together:
Total Resistance (RT) = 3Ω + 4Ω + 6Ω
RT = 13Ω
To maintain a current of 2A through the circuit, use Ohm's Law to calculate the required voltage:
Voltage (V) = Current (I) × Resistance (R)
V = 2A × 13Ω
V = 26V
Since the power source has a voltage of 12V, you cannot achieve a current of 2A by arranging the resistors in series.
However, if you have made an error in the problem statement or if there is another power source with a voltage of 26V, the resistors should be arranged in parallel to achieve the desired current. In a parallel circuit, the total resistance is given by the reciprocal of the sum of the reciprocals of the individual resistances:
1/RT = 1/3Ω + 1/4Ω + 1/6Ω
1/RT = (8 + 6 + 4)/24
1/RT = 18/24
RT = 24/18
RT = 4/3 Ω
Now, you can use Ohm's Law to calculate the required voltage:
Voltage (V) = Current (I) × Resistance (R)
12V = 2A × (4/3)Ω
12V = 8/3V
In this case, adjusting the resistors in parallel will not achieve the desired current of 2A.