Series Circuit Maximum Voltage of a Resistor

[Inquisitive]

So, I'm refreshing my understanding and working through a textbook of problems.

I'm not getting the same answer as the textbook. Here is the problem.

A 1-kΩ ¼ W resistor and a 1.5-kΩ 1/8 W resistor are in series. What is the maximum amount of voltage that can be applied to this circuit without exceeding the wattage rating of either resistor?

My calculations are:

V = sqrt(0.25 x 1000)

V1 = 15.811

V = sqrt(0.125 x 1500)

V2 = 13.693

Vmax = 15.811 + 13.693

Vmax = 29.504 is my answer

The textbook says Vmax = 22.82V ???

Where did I go wrong? Did I overlook something?

 

[audioguru]

You calculated the maximum power in each resistor separately so their currents were different. But in the circuit the resistors are in series so the current in each one must be the same.
Calculate the maximum power of each resistor then calculate its current. Then with each resistor using the current that is the smallest, calculate the total voltage.

It is a silly exercise because NOBODY uses a resistor at its maximum allowed power rating when it is almost red hot, it might desolder itself, it will melt any plastic capacitor or case nearby and will be even hotter when enclosed/and or in summer.

 

[Inquisitive]

Did try that also and got 11.18 + 13.693 = 24.873V. Based on 0.125 current in both resistors.

This is still 2.053V too high of the textbook answer.

 

[JimB]

I get 22.8 volts.

Calculate the current for rated disipation in each individual resistor.
Take the lower of the two currents, and multiply by the total resistance.
You should get 22.8 or there abouts.

JimB

 

[Ratchit]

Did try that also and got 11.18 + 13.693 = 24.873V. Based on 0.125 current in both resistors.

This is still 2.053V too high of the textbook answer.

Your textbook is correct. First, find the maximum current that each resistor can pass at their rated value. Then use the smaller of the maximum current values and calculate the voltage across both resistors when that current exists in series with the two resistors. If that current value is exceeded, one of the resistors will exceed its max rating.

Ratch

 

[Inquisitive]

Got it. But nobody would push a component that hard.

 

[Inquisitive]

Thus, V = √P x R
√(0.125 x 1500) = 13.693V (R1 Voltage)

I = V/R
13.693/1500 = 0.009128A or 9.128 mA

V = IR
0.009128A x 1000Ω = 9.128 V (R2 Voltage)

13.693V + 9.128V = 22.821 V or 22.82V

Thanks for the help gentlemen.