Consider the circuit shown in the figure below. (Assume R1 = 12.0 Ω and R2 = 6.00 Ω.)
Read this for the figure -
A rectangular circuit begins at its left side at a 25.0 V battery with the positive terminal on top. From the battery, the circuit extends up and to the right to point b. The circuit continues rightward from point b and splits into three parallel vertical branches. The leftmost branch has a resistor labeled R1, the middle branch has a resistor labeled R2 and the rightmost branch, from top to bottom, goes through a resistor labeled R2, point c, then and a resistor labeled 20.0 Ω. The three branches recombine, the circuit extends to the left passing through point a, bends upward to a resistor labeled R1 and reaches the negative terminal of the battery.
(a) Find the potential difference between points a and b.
(b) Find the current in the 20.0-Ω resistor.
Over on the right you have three branches
in ohms
left 12
middle 6
right 6 + 20 = 26
find Re or equivalent resistance of those three right legs
1/Re = 1/12 + 1/6 + 1/26 = .08333+ .16666 + .03846 = 0.28845
so Re = 3.467 ohms
resistance around the circuit = 3.467 + 12 = 15.47 ohms
so total current = 25/15.47 = 1.61 amps
so voltage drop across bottom R1 = 1.61 * 12 = 19.4 volts
so voltage drop across Re = 25 -19.4 = 5.61 volts (that is answer 1)
so current through rightmost leg = 5.61 / 26 = .0235 amps
(a) Well, the potential difference between points a and b really depends on how many dad jokes you've told recently. Just kidding! Let's get to the serious part. To find the potential difference between a and b, we need to add up the potential drops along the circuit.
Starting from the battery, we have a potential difference of 25.0V. Moving from left to right, we encounter resistor R1 first. The potential drop across R1 is given by Ohm's Law: V = IR, where V is the potential drop, I is the current, and R is the resistance. Plugging in the values, we have V = (I1)(R1) = (I1)(12.0Ω).
Next, in the middle branch, we have resistor R2. Following the same logic, the potential drop across R2 is V = (I2)(R2) = (I2)(6.00Ω).
Finally, on the rightmost branch, we have two resistors in series - R2 and the 20.0Ω resistor. Since they are in series, the current passing through both resistors is the same. Let's call this current I3. Therefore, the potential drop across these two resistors is V = (I3)(R2 + 20.0Ω).
Now, since the three branches merge together again, the sum of the potential drops across the individual resistors must be equal to the potential difference across points a and b.
So, the equation becomes:
25.0V = (I1)(12.0Ω) + (I2)(6.00Ω) + (I3)(R2 + 20.0Ω)
Now, it's up to you to solve this equation and find the potential difference between points a and b!
(b) As for the current in the 20.0-Ω resistor, we can use Ohm's Law once again. The potential difference across the 20.0-Ω resistor is already given by the equation in part (a), and we know that the resistance is 20.0Ω. Therefore, we can use the equation V = IR to solve for I:
I = V / R
Plug in the numbers and calculate the current flowing through the 20.0-Ω resistor!
Remember, while this circuit may seem confusing, just remember that laughter is the best conductor of electricity!
To find the potential difference between points a and b (Va - Vb), we can use Kirchhoff's voltage law, which states that the sum of the potential differences around any closed loop in a circuit is zero.
Let's start by analyzing the circuit step by step:
Step 1: Calculate the equivalent resistance of the parallel branches.
- The resistance of the first parallel branch (R1) is given as 12.0 Ω.
- The resistance of the second parallel branch (R2) is given as 6.00 Ω.
- The resistance of the third parallel branch is given as the sum of R2 and 20.0 Ω, which is 26.0 Ω (6.00 Ω + 20.0 Ω).
- To find the equivalent resistance of the parallel branches (Rp), we can use the formula:
1/Rp = 1/R1 + 1/R2 + 1/R3
1/Rp = 1/12.0 Ω + 1/6.00 Ω + 1/26.0 Ω
1/Rp = (2/24.0 + 4/24.0 + 1/26.0) Ω^-1
1/Rp = 7/24.0 + 1/26.0 Ω^-1
1/Rp = (182 + 24)/24 * 26 Ω^-1
1/Rp = 206/624 Ω^-1
1/Rp = 1/3.03 Ω^-1
Rp = 3.03 Ω (approximately)
Step 2: Calculate the total resistance of the circuit.
- The total resistance of the circuit (Rt) is given by the sum of the resistances in series along the circuit path.
- Along the path from a to b, we have R1, Rp, and R1 again in series.
- Rt = R1 + Rp + R1
- Rt = 12.0 Ω + 3.03 Ω + 12.0 Ω
- Rt = 27.03 Ω (approximately)
Step 3: Calculate the potential difference between points a and b (Va - Vb).
- To find Va - Vb, we can use Ohm's law:
Va - Vb = I * Rt
- Let's assume the current flowing through the circuit is I.
- We need to find the value of I to calculate Va - Vb.
Step 4: Calculate the current in the circuit.
- To find the current flowing through the circuit, we can again use Ohm's law.
- Along the path from a to b, we have a 25.0 V battery supplying energy.
- Thus, Va - Vb = 25.0 V.
- Using Ohm's law, I can be found as:
I = (Va - Vb) / Rt
I = 25.0 V / 27.03 Ω
I ≈ 0.925 A (approximately)
Step 5: Calculate Va - Vb.
- Now that we have the value of I, we can calculate Va - Vb.
- Va - Vb = I * Rt
- Va - Vb = 0.925 A * 27.03 Ω
- Va - Vb ≈ 25.0 V (approximately)
(a) The potential difference between points a and b is approximately 25.0 V.
Step 6: Calculate the current in the 20.0-Ω resistor.
- The current flowing through the 20.0-Ω resistor is the same as the total current in the circuit, which we found to be 0.925 A.
(b) The current in the 20.0-Ω resistor is approximately 0.925 A.
To solve this circuit, we need to apply the principles of Ohm's Law and Kirchhoff's Rules.
(a) To find the potential difference between points a and b, we can use Kirchhoff's Voltage Law (KVL), which states that the sum of the voltage drops in a closed loop is zero.
Starting from point a and moving clockwise around the loop:
1. From point a to point b, we have a resistor R1 with resistance 12.0 Ω. The voltage drop across this resistor can be calculated using Ohm's Law: V1 = I * R1, where I is the current flowing through R1. Let's call this V1.
2. From point b to point c, we have a resistor R2 with resistance 6.00 Ω. The voltage drop across this resistor can be calculated similarly: V2 = I * R2.
3. From point c to point a, we have two resistors in series: R2 with resistance 6.00 Ω and the 20.0 Ω resistor. The total resistance is R_total = R2 + 20.0 Ω. The voltage drop across these resistors can be calculated: V3 = I * R_total.
Now, applying KVL:
V1 - V2 - V3 = 0
Substituting the values:
I * R1 - I * R2 - I * R_total = 0
I * (R1 - R2 - R_total) = 0
I = 0, which means that there is no current flowing through the circuit. Therefore, the potential difference between points a and b is 0 V.
(b) Since there is no current flowing through the circuit, the current in the 20.0-Ω resistor is also 0 A.