What is the equivalent resistance of a parallel circuit with a 5.0 kΩ resistor, a 2.5 kΩ resistor, and 10.0 kΩ resistor?
A. 0.10 kΩ
b. 0.83 kΩ
c. 10 kΩ
d. 1.4 kΩ
e. 18 kΩ
Req
=1/(1/R1+1/R2+1/R3+...)
So, with 3 resistors,
Req
=1/(1/5+1/2.5+1/10)
=...
1/R = 1/5 + 2/5 + 1/10 = (2+4+1)/10 = 7/10
so R = (10/7) k Ohms =1.44 k Ohms
thx, MathMate and Damon :)
Well, in a parallel circuit, it's like a group of friends trying to make a decision together. They can't make up their minds, so they split up and go their own ways. Similarly, the current splits up and flows through each resistor in a parallel circuit.
To find the equivalent resistance in a parallel circuit, you can use the formula 1/R_total = 1/R1 + 1/R2 + 1/R3 + …, where R_total is the equivalent resistance and R1, R2, R3, etc. are the individual resistances.
So, let's plug in the values and do the math:
1/R_total = 1/5.0 + 1/2.5 + 1/10
To get the equivalent resistance, we take the reciprocal of both sides:
R_total = 1/(1/5.0 + 1/2.5 + 1/10)
Now, time to simplify:
R_total = 1/(2/10 + 4/10 + 1/10)
R_total = 1/(7/10)
R_total = 10/7
And voila, the equivalent resistance is approximately 1.4 kΩ. So, the answer is (d) 1.4 kΩ.
Remember, resistance may slow down the flow of current, but it can't resist the puns!
To find the equivalent resistance of a parallel circuit, you can use the formula:
1/Req = 1/R1 + 1/R2 + 1/R3 + ...
In this case, the resistors are 5.0 kΩ, 2.5 kΩ, and 10.0 kΩ. To calculate the equivalent resistance, substitute these values into the formula:
1/Req = 1/5.0 kΩ + 1/2.5 kΩ + 1/10.0 kΩ
To simplify this equation, you need to find a common denominator:
1/Req = (2/10.0 kΩ) + (4/10.0 kΩ) + (1/10.0 kΩ)
1/Req = (2 + 4 + 1)/10.0 kΩ
1/Req = 7/10.0 kΩ
To get the equivalent resistance (Req), you take the reciprocal of both sides of the equation:
Req = 10.0 kΩ / 7
Now, divide 10.0 kΩ by 7:
Req ≈ 1.43 kΩ
So, the closest option to the calculated equivalent resistance is:
d. 1.4 kΩ