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. = The answer is 5.61 volts.
(b) Find the current in the 20.0-Ω resistor. = ??????
To find the current in the 20.0-Ω resistor, we can use Ohm's Law and the concept of parallel resistors.
Let's label the current in the 20.0-Ω resistor as I_20 and the total current flowing through the circuit as I_total.
In a parallel circuit, the total current is equal to the sum of the currents in the individual branches. So we can write:
I_total = I_20 + I_R1 + I_R2
Now let's focus on finding the currents in the resistors R1 and R2.
For R1, we can use Ohm's Law:
I_R1 = V_R1 / R1
To find V_R1, we need to find the potential difference across R1. We know that the potential difference across points a and b is 5.61 volts (as given in part a). Therefore, V_R1 = 5.61 V.
For R2, we can also use Ohm's Law:
I_R2 = V_R2 / R2
To find V_R2, we need to find the potential difference across R2. From the figure, we see that the potential difference across points b and c is the same as the potential difference across points a and b. Therefore, V_R2 = 5.61 V.
Now let's substitute these values back into the equation for total current:
I_total = I_20 + (V_R1 / R1) + (V_R2 / R2)
We know that the potential difference across the battery is 25.0 V, and it is connected to a parallel circuit. So the potential difference across each branch is equal to 25.0 V.
Therefore, V_R2 + V_c = 25.0 V
Since V_R2 = 5.61 V, we can find V_c by subtracting 5.61 V from 25.0 V:
V_c = 25.0 V - 5.61 V = 19.39 V
Now, we can determine the current I_20 using Ohm's Law:
I_20 = V_c / 20.0 Ω
Substituting the value of V_c, we find:
I_20 = 19.39 V / 20.0 Ω = 0.97 A
Therefore, the current in the 20.0-Ω resistor is 0.97 amps.