You have several 36 ohm resistors and a 12 Volt battery.
d. When you connect one resistor to the battery, what is the current of the circuit?
e. When you wire two resistors in series to the battery, what is the current of the circuit?
f. When you wire two resistors in parallel to the battery, what is the current of the circuit?
i = V/R
d. V = 12 , R = 36
e, V = 12 , R = 72
f . V = 12 , 1/R = 1/36 + 1/36 = 2/36 = 1/18 so R = 18 ohms
( twice as much current as in part d because two equal routes for the currnet to use.)
d. I = E/R = 12/36 = __ Amps.
e. I = E/(R1+R2) = 12/72 =
f. 1/R = 1/R1 + 1/R2 = 1/36 + 1/36 = 2/36 = 1/18
R = 18 ohms.
I = E/R =
To find the current of the circuit, we'll need to use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the total resistance (R).
Given information:
Resistance of one resistor (R1) = 36 ohms
Voltage of the battery (V) = 12 Volts
d. When you connect one resistor to the battery:
Since there is only one resistor connected, the total resistance (R_total) in the circuit is equal to the resistance of that one resistor.
R_total = R1 = 36 ohms
Using Ohm's Law:
I = V / R_total
I = 12 V / 36 Ω
I ≈ 0.33 Amps
Therefore, the current of the circuit when one resistor is connected to the battery is approximately 0.33 Amps.
e. When you wire two resistors in series to the battery:
When resistors are wired in series, their resistances add up to give the total resistance (R_total) of the circuit.
R_total = R1 + R2
Since we have two resistors with the same resistance:
R_total = R1 + R2 = 36 Ω + 36 Ω = 72 Ω
Using Ohm's Law:
I = V / R_total
I = 12 V / 72 Ω
I ≈ 0.17 Amps
Therefore, the current of the circuit when two resistors are wired in series to the battery is approximately 0.17 Amps.
f. When you wire two resistors in parallel to the battery:
When resistors are wired in parallel, the total resistance (R_total) is calculated differently.
R_total = (1 / R1) + (1 / R2)
Since we have two resistors with the same resistance:
R_total = (1 / R1) + (1 / R2) = (1 / 36 Ω) + (1 / 36 Ω) = 2 / 36 Ω
R_total = 1 / (2 / 36) = 1 / (1 / 18) = 18 Ω
Using Ohm's Law:
I = V / R_total
I = 12 V / 18 Ω
I ≈ 0.67 Amps
Therefore, the current of the circuit when two resistors are wired in parallel to the battery is approximately 0.67 Amps.
To find the current of a circuit, you can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R): I = V / R.
d. When you connect one resistor to the battery, the current of the circuit can be calculated by dividing the voltage of the battery (12V) by the resistance of one resistor (36Ω):
I = 12V / 36Ω = 0.33 Amperes (A) or 330 milliamperes (mA).
e. When you wire two resistors in series to the battery, the total resistance (R_total) of the circuit is the sum of the resistances of the individual resistors. In this case, the total resistance would be 36Ω + 36Ω = 72Ω. Using Ohm's Law, you can find the current:
I = 12V / 72Ω = 0.17A or 170mA.
f. When you wire two resistors in parallel to the battery, the total resistance (R_total) of the circuit can be calculated using the formula: 1/R_total = 1/R1 + 1/R2. In this case, the total resistance would be 1/36Ω + 1/36Ω = 1/18Ω. Taking the reciprocal of this gives us the total resistance: R_total = 18Ω. Now, we can find the current using Ohm's Law:
I = 12V / 18Ω = 0.67A or 670mA.
So, to summarize:
d. When you connect one resistor to the battery, the current of the circuit is 0.33A or 330mA.
e. When you wire two resistors in series to the battery, the current of the circuit is 0.17A or 170mA.
f. When you wire two resistors in parallel to the battery, the current of the circuit is 0.67A or 670mA.