A direct current power supply is connected to a circuit containing one resistor, and 1.5 A is drawn. If a second identical resistor is connected to the power supply in parallel with the first, how much current is drawn from the battery?
resistance is halved, so current is doubled.
To find out how much current is drawn from the battery when a second identical resistor is connected in parallel, you can use the concept of electrical resistance in parallel circuits and apply Ohm's Law.
Ohm's Law states that the current flowing through a conductor (in this case, a resistor) is equal to the voltage across the conductor divided by its resistance:
I = V / R
In the given scenario, we know the following:
- The current drawn from the battery when only one resistor is connected is 1.5 A.
Since the two resistors are identical and connected in parallel, the total resistance in the circuit will be different. In a parallel configuration, the total resistance is calculated as the reciprocal of the sum of the reciprocals of the individual resistances:
1/R_total = (1/R1) + (1/R2)
Since both resistors are identical, we can simplify this equation to:
1/R_total = (1/R) + (1/R)
1/R_total = 2/R
Now, we can find the total resistance (R_total) in terms of the given resistance (R):
R_total = R/2
To find the current (I_total) drawn from the battery when the second resistor is added in parallel, we can rearrange Ohm's Law:
I_total = V / R_total
Since the power supply is a direct current (DC) power supply, the voltage across each resistor in the circuit will be the same. Therefore, we can substitute V with the same value and rewrite the equation as:
I_total = V / (R/2)
Since the current drawn when only one resistor is connected is 1.5 A, we can substitute the values into the equation:
1.5 A = V / (R/2)
To find the value of I_total, we need to know the value of R. Once the value of R is known, we can rearrange the equation to solve for I_total.