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?

Answer 1

For series,

n = 4 resistors

For parallel

n = 4 resistors

Step-by-step 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