11. If you have a circuit set up where you have have a 9 V battery and one resistor of 10 Ohms. You place a voltmeter across the two ends of the battery and it reads 8.333 V. What is the internal resistance of the battery?
12. A 12 V battery is set up in an unknown configuration. The internal resistance of the battery is 4 Ohms. If this circuit must have a minimum of two resistors, what resistance must these resistors have and what configuration must they be in in order for a terminal voltage to read 11.478 V? Show all work
11. Assuming the no-load voltage is 9 volts:
I = V/R = 8.333/10 = 0.833A.
r = (E-Vl)/I = (9-8.333)/0.833 =
12. Given: E = 12V., r = 4 ohms, Vl = 11.478V.
I = (E-Vl)/r = (12-11.478)/4 = 0.131A.
Req = V/I = 11.478/0.131 = 88 ohms. = R1 and R2 in-parallel.
R1 = R2 = 2*88 = 176 ohms each.
How did you know its in a parallel? I believe that you are right btw.
Series connection: R1 = R2 = 88/2 = 44 ohms.
To answer question 11, we need to use Kirchhoff's voltage law (KVL) and Ohm's law. Kirchhoff's voltage law states that the sum of the voltages in a closed loop in a circuit is zero. Ohm's law states that the voltage across a resistor is equal to the current through it multiplied by its resistance.
In this case, we can consider the internal resistance of the battery, denoted as r (in Ohms). The voltage across the resistor, Vr (in volts), can be calculated using Ohm's law as follows:
Vr = (9 V - 8.333 V)
Since the voltage across the resistor is equal to the voltage across the battery minus the voltage measured by the voltmeter, we have:
Vr = (9 V - 8.333 V) = 0.667 V
Now, applying Kirchhoff's voltage law to the circuit, we can write:
9 V - Vr - r * I = 0
Where I represents the current flowing through the circuit. Since we only have one resistor in the circuit, this current is equal to the current flowing through that resistor, which can be calculated using Ohm's law:
I = Vr / R
We know the value of Vr (0.667 V) and R (10 Ohms), so we can substitute these values into the equation to solve for I:
I = 0.667 V / 10 Ohms = 0.0667 A
Now, substituting the calculated value of I back into the equation from Kirchhoff's voltage law, we can solve for the internal resistance of the battery, r:
9 V - 0.667 V - r * 0.0667 A = 0
Simplifying the equation further:
9 V - 0.0677 V - 0.0667 A * r = 0
8.3323 V - 0.0667 A * r = 0
0.0667 A * r = 8.3323 V
r = 8.3323 V / 0.0667 A
r ≈ 124.87 Ohms
So, the internal resistance of the battery is approximately 124.87 Ohms.
Now let's move on to question 12. We need to determine the resistance of the two unknown resistors and their configuration in order to achieve a terminal voltage of 11.478 V using a 12 V battery with an internal resistance of 4 Ohms.
Let's assume the two resistors have values R1 and R2 (in Ohms). Since the resistors are in unknown configuration, we'll consider two possible setups:
Option 1: The resistors are in series configuration.
In a series configuration, the total resistance (Rt) is the sum of individual resistances:
Rt = R1 + R2
Applying Ohm's law to this configuration, the terminal voltage can be calculated:
Vt = I * Rt
where I is the current flowing through the circuit. Since the circuit involves only resistors and a battery, the current can be determined using Ohm's law:
I = (12 V - Vt) / (4 Ohms + R1 + R2)
Substituting this expression for I into the formula for Vt, we have:
Vt = ((12 V - Vt) / (4 Ohms + R1 + R2)) * (R1 + R2)
Simplifying this equation, we get:
Vt = (12 V - Vt) * (R1 + R2) / (4 Ohms + R1 + R2)
Now, let's solve for Vt:
Vt = (12 V - Vt) * (R1 + R2) / (4 Ohms + R1 + R2)
Multiplying both sides of the equation by (4 Ohms + R1 + R2) to eliminate the denominator:
Vt * (4 Ohms + R1 + R2) = (12 V - Vt) * (R1 + R2)
4 Vt Ohms + Vt * R1 + Vt * R2 = 12 V * R1 + 12 V * R2 - Vt * R1 - Vt * R2
4 Vt Ohms = 12 V * R1 + 12 V * R2 - Vt * R1 - Vt * R2
Simplifying further:
4 Vt Ohms = (12 V - Vt) * (R1 + R2)
Dividing both sides of the equation by (12 V - Vt), we can solve for the sum of the resistances, (R1 + R2):
4 Vt Ohms / (12 V - Vt) = R1 + R2
Now, let's substitute the given value of the terminal voltage (11.478 V) into the equation and solve for R1 + R2:
4 * 11.478 V / (12 V - 11.478 V) = R1 + R2
45.912 V / 0.522 V = R1 + R2
88.013 Ohms = R1 + R2
So, the sum of the resistances R1 and R2 must be approximately 88.013 Ohms in order to achieve a terminal voltage of 11.478 V.
We have found the necessary resistance configuration for the two unknown resistors. However, we do not have enough information to determine the individual resistance values of R1 and R2. Further information or equations are required to solve for those values.