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
This is all I understand. Can you explain step by step and show how to get the answer?
1/R = 1/Ra + 1/Rb + 1/Rc +1/Rd
? = 1/10.0 + 1/10.0 + 1/10.0 + 1/10.0
well
1/Req= 4/10 or Req=2.5 ohms.
Now, add the 40 ohms in series, so Rtotal=42.5ohms
To find the equivalent resistance of the circuit, we need to calculate the total resistance of the parallel arrangement and then add it to the resistance of the series arrangement.
Step 1: Calculate the total resistance of the parallel arrangement.
The formula for calculating the total resistance of resistors connected in parallel is:
1/R = 1/Ra + 1/Rb + 1/Rc + 1/Rd
In this case, all resistors have a resistance of 10.0 Ω, so we can substitute the values into the formula:
1/R = 1/10.0 + 1/10.0 + 1/10.0 + 1/10.0
Step 2: Simplify the equation.
To simplify the equation, we can add the fractions together:
1/R = 0.1 + 0.1 + 0.1 + 0.1
1/R = 0.4
Step 3: Solve for R.
To find the equivalent resistance (R), we need to take the reciprocal of both sides of the equation:
R = 1/0.4
R = 2.5 Ω
Step 4: Calculate the total resistance of the series arrangement.
In the series arrangement, all four resistors are connected in series, so their resistances add up:
Total resistance of series arrangement = 4 * 10.0 Ω = 40.0 Ω
Step 5: Add the total resistance of the parallel arrangement and the series arrangement together to find the equivalent resistance of the circuit:
Equivalent resistance = Total resistance of parallel arrangement + Total resistance of series arrangement
= 2.5 Ω + 40.0 Ω
= 42.5 Ω
Therefore, the equivalent resistance of the circuit is 42.5 Ω.
The correct answer is option D) 42.5.