1)Four resistors of 10.0 each are connected in parallel. A combination of four resistors of 10.0 each is connected in series along with the parallel arrangement. What is the equivalent resistance of the circuit?
A)80.0
B)40.4
C)40.0
D)42.5
I got C
2)A 2.00 resistor, and a 12.0 resistor are connected in parallel across a 20.0-V battery. What is the current flowing through the 2.00 resistor?
A)1.67 A
B)1.40 A
C)11.6 A
D)10.0 A
C=vsource/R
I got D
1 is wrong.
2 is wrong. I don't know what you used for R.
1)resistance in parallel = 40 resistance in series = 40
total equaivalent resistance = 80 which is A
2)C = vsource/R
C = 20.0/2.00
= D. How could it be anything else
1) Since when do parallel resistors add the resistances?
2) The two ohm reistor is in parallel with the twelve ohm resistance. Total resistance then of the parallel branch will be LESS than two ohm.
You need to review what happens to resistance when resistors are connected in parallel.
To solve these circuit problems, you need to understand the concepts of parallel and series connections.
1) For the first question, you are given four resistors of 10.0 Ω each connected in parallel. To find the equivalent resistance of the parallel arrangement, you can use the formula:
1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4
Substituting the given values:
1/Req = 1/10.0 + 1/10.0 + 1/10.0 + 1/10.0
Simplifying this equation gives:
1/Req = 4/10.0
1/Req = 0.4
Taking the reciprocal of both sides:
Req = 1/0.4
Req = 2.5 Ω
Now, this parallel arrangement is connected in series along with another combination of four resistors of 10.0 Ω each. Since resistors connected in series simply add up, you can add the equivalent resistance of the parallel arrangement with the resistance of the series arrangement:
Req_total = Req_parallel + Req_series
Req_total = 2.5 + 10.0 + 10.0 + 10.0 + 10.0
Req_total = 42.5 Ω
Therefore, the equivalent resistance of the circuit is 42.5 Ω, which corresponds to answer choice D.
2) For the second question, the resistors of 2.00 Ω and 12.0 Ω are connected in parallel. To find the current flowing through the 2.00 Ω resistor, you can use Ohm's Law:
I = V/R
Substituting the given values:
I = 20.0 V / 2.00 Ω
I = 10.0 A
Therefore, the current flowing through the 2.00 Ω resistor is 10.0 A, which corresponds to answer choice D.