A 50 V battery has an internal resistance r.
1 A
50 V
r
55 Ω
66 Ω
internal
resistance
What is the value of r?
Answer in units of Ω
009 (part 2 of 2) 10.0 points
Determine the magnitude of the current
through the 66 Ω resistor in the upper left
of the circuit.
Answer in units of A
To find the value of internal resistance r, we can use Ohm's Law. Ohm's Law states that the current flowing through a circuit is equal to the voltage divided by the total resistance.
In this case, we have a 50 V battery with an internal resistance r. The total resistance is the sum of the internal resistance and the 66 Ω resistor, which is 66 Ω + r.
Therefore, the current flowing through the circuit can be calculated using the formula:
Current = Voltage / Total Resistance
Now, let's substitute in the given values to find the value of r.
Current = 1 A
Voltage = 50 V
Total Resistance = 66 Ω + r
Substituting these values into the formula, we get:
1 A = 50 V / (66 Ω + r)
To find the value of r, we can solve this equation for r. Rearranging the equation, we get:
66 Ω + r = 50 V / 1 A
66 Ω + r = 50 Ω
Now, we subtract 66 Ω from both sides of the equation:
r = 50 Ω - 66 Ω
r = -16 Ω
Therefore, the value of r is -16 Ω.
However, it is important to note that negative internal resistance is not physically meaningful. Internal resistance is typically positive, so this result may indicate a mistake or inconsistency in the given values.
Now, let's move on to determining the magnitude of the current through the 66 Ω resistor.
The magnitude of the current through a resistor can be calculated using Ohm's Law. Again, Ohm's Law states that the current flowing through a circuit is equal to the voltage divided by the resistance.
In this case, the 66 Ω resistor is connected directly to the battery, so the voltage across it is equal to the battery voltage, which is 50 V.
To find the current, we can use the formula:
Current = Voltage / Resistance
Substituting the given values, we get:
Current = 50 V / 66 Ω
Calculating this, we find:
Current ≈ 0.758 A
Therefore, the magnitude of the current through the 66 Ω resistor is approximately 0.758 A.