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You are given a number of 12 Ω resistors, each capable of dissipating only 2.6 W without being destroyed. What is the minimum number of such
Question
You are given a number of 12 Ω resistors, each capable of dissipating only 2.6 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 12 Ω resistance that is capable of dissipating at least 10.1 W?
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2021-09-05T10:18:46+00:00
2021-09-05T10:18:46+00:00 1 Answers
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Answer:
For series,
n = 4 resistors
For parallel
n = 4 resistors
Explanation:
From the question,
For Series
Note: When resistors are connected in series, the same current flows through each of the resistors and the combined resistance
If 12Ω resistor each is capable of dissipating 2.6 W.
P = I²R…………….. Equation 1
Where P = power, R = resistance of the resistor, I = current.
make I the subject of the equation
I = √(P/R)…………. Equation 2
Given: P = 2.6 W, R = 12 Ω
Substitute into equation 2
I = √(2.6/12)
I = 0.4655 A
The combined number of 12 Ω resistor in series is given as
R’ = Rn…………….. Equation 3
Where n = minimum number of resistor, R’ = combined resistance
R’ = 12n
10.1 = (0.4655)²12n
10.1 = 0.2167×12n
10.1 = 2.6n
n = 10.1/2.6
n = 3.88
n ≈ 4 resistor
For parallel,
Note: when resistors are connected in parallel, The have the same voltage with the combined resistance
P = V²/R………………… Equation 4
make V the subject of the equation
V = √(PR)
V = √(2.6×12)
V = √31.2 V
The combined resistance for parallel connection
R’ = R/n
R’ = 12/n
Substitute into equation 4
10.1= (√31.2)²n/12
10.1×12 = 31.2n
121.2 = 31.2n
n = 121.2/31.2
n = 3.88
n = 4 resistors