Consider the circuit shown in the figure below. Use the following as necessary: R1 = 11.00 Ù, R2 = 1.40 Ù, and V = 6.00 V.
(a) Calculate the equivalent resistance of the R1 and 5.00 Ù resistors connected in parallel.
1 Ù
(b) Using the result of part (a), calculate the combined resistance of the R1, 5.00 Ù, and 4.00 Ù resistors.
2 Ù
(c) Calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00 Ù resistor.
3 Ù
(d) Combine the equivalent resistance found in part (c) with the R2 resistor.
4 Ù
(e) Calculate the total current in the circuit.
5 A
(f) What is the voltage drop across the R2 resistor?
6 V
(g) Subtracting the result of part (f) from the battery voltage, find the voltage across the 3.00 Ù resistor.
7 V
(h) Calculate the current in the 3.00 Ù resistor.
8 A
Sorry, without the accompanying figure, it is quite difficult to guess what the circuit is like.
See if you can post a link, although it is not always possible.
To solve this circuit problem, we need to apply the principles of series and parallel resistors. Let's break down each part and explain how to get the answer.
(a) To calculate the equivalent resistance of R1 and the 5.00 Ù resistor connected in parallel, we use the formula for the total resistance of parallel resistors:
1/Req = 1/R1 + 1/Rparallel
Given R1 = 11.00 Ù and Rparallel = 5.00 Ù, we substitute the values into the formula:
1/Req = 1/11.00 + 1/5.00
Simplifying this expression, we get:
1/Req = (5 + 11) / (11 * 5)
1/Req = 16 / 55
Taking the reciprocal of both sides:
Req = 55 / 16
Therefore, the equivalent resistance of R1 and the 5.00 Ù resistor connected in parallel is approximately 3.4375 Ù.
(b) Using the result of part (a), we calculate the combined resistance of R1, the 5.00 Ù resistor, and the 4.00 Ù resistor. Since R1 and the 5.00 Ù resistor are in parallel, we can treat them as a single equivalent resistor Req_parallel. Then, we have two resistors, Req_parallel and the 4.00 Ù resistor, connected in series. Thus, to find the combined resistance, we simply add the resistances:
Rcombined = Req_parallel + R2
Given Req_parallel = 3.4375 Ù and R2 = 1.40 Ù, we substitute the values into the formula:
Rcombined = 3.4375 + 1.40
Therefore, the combined resistance of R1, the 5.00 Ù resistor, and the 4.00 Ù resistor is approximately 4.8375 Ù.
(c) To calculate the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00 Ù resistor, we again use the formula for the total resistance of parallel resistors:
1/Req = 1/Rcombined + 1/Rparallel
Given Rcombined = 4.8375 Ù and Rparallel = 3.00 Ù, we substitute the values into the formula:
1/Req = 1/4.8375 + 1/3.00
Simplifying this expression, we get:
1/Req = (3.00 + 4.8375) / (4.8375 * 3.00)
1/Req = 7.8375 / 14.5125
Taking the reciprocal of both sides:
Req = 14.5125 / 7.8375
Therefore, the equivalent resistance of the combined resistance found in part (b) and the parallel 3.00 Ù resistor is approximately 1.853 Ù.
(d) Combining the equivalent resistance found in part (c) with the R2 resistor, which are in series, the total resistance R_total is given by:
R_total = Req + R2
Given Req = 1.853 Ù and R2 = 1.40 Ù, we substitute the values into the formula:
R_total = 1.853 + 1.40
Therefore, the total resistance in the circuit is approximately 3.253 Ù.
(e) To calculate the total current in the circuit, we use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R):
I = V / R_total
Given V = 6.00 V and R_total = 3.253 Ù, we substitute the values into the formula:
I = 6.00 / 3.253
Therefore, the total current in the circuit is approximately 1.846 A.
(f) To find the voltage drop across the R2 resistor, we use Ohm's Law again:
V_drop_R2 = I * R2
Given I = 1.846 A and R2 = 1.40 Ù, we substitute the values into the formula:
V_drop_R2 = 1.846 * 1.40
Therefore, the voltage drop across the R2 resistor is approximately 2.584 V.
(g) To find the voltage across the 3.00 Ù resistor, we subtract the result of part (f) from the battery voltage (V):
V_3Ù = V - V_drop_R2
Given V = 6.00 V and V_drop_R2 = 2.584 V, we substitute the values into the formula:
V_3Ù = 6.00 - 2.584
Therefore, the voltage across the 3.00 Ù resistor is approximately 3.416 V.
(h) To calculate the current in the 3.00 Ù resistor, we use Ohm's Law:
I_3Ù = V_3Ù / R3
Given V_3Ù = 3.416 V and R3 = 3.00 Ù, we substitute the values into the formula:
I_3Ù = 3.416 / 3.00
Therefore, the current in the 3.00 Ù resistor is approximately 1.138 A.