Calculating Maximum Permissible Current

  • Thread starter Thread starter licklecee
  • Start date Start date
  • Tags Tags
    Current Maximum
AI Thread Summary
The discussion focuses on calculating the maximum permissible current and voltage for a 1-watt resistor with a resistance of 10,000 ohms, factoring in a 5% tolerance. The initial calculation of current yielded 0.01 mA, but participants clarified that the maximum current should be calculated using the maximum resistance of 10,500 ohms, resulting in approximately 9.76 mA. Additionally, to find the maximum voltage, the lowest resistance value of 9,500 ohms was used, leading to a voltage of about 92.72 volts. Participants emphasized the importance of correctly applying the resistor's tolerance to determine accurate values. Overall, the calculations highlight the need for careful attention to detail in electrical calculations.
licklecee
Messages
8
Reaction score
0

Homework Statement



B) What is the maximum permissible current which can flow through a 1 watt resistor colour coded Brown, Black Orange, Gold ?

C ) what is the highest voltage that can be applied to the resistor in part ( B ) ?


Homework Equations



P = I^2 * R



The Attempt at a Solution



P = 1w
R = 10,000

I = sqrt(P/R)
I = sqrt(1/10,000)
I = sqrt 0.00001
I = 0.01 Ma

Variance Factor :

Gold = 5%

0.01 * 1.05 = 0.0105Ma

Maximum Current = 0.0105Ma

Can someone please double check this for me ? I'm struggling abit with this problem. Thank you!
 
Physics news on Phys.org
1. Recheck sqrt(1/10000) = sqrt(0.00001).

2. Why are you expressing current in mA?

3. Yes the tolerance is 5% but that should not be used to calculate maximum current. (The 5% refers to the resistance, so the resistance can vary from 9500 ohms to 10500 ohms. This means that in order to dissipate 1 W, the 9500 ohm resistor would require sqrt(1/9500) amps or 10.26 mA and the 10500 ohm resistor would require sqrt(1/10500) amps or 9.76 mA.) Note that a 5% change in resistance results in approximately 2.5% change in current.
 
oh okay,

1.so sqrt 0.00001 = 3.16

2. Would it be 3.16 amps then ?

3. Okay so is it correct to say that to calculate maximum current, i need to calculate the maximum resistance, using the tolerance of in this case 5 percent,. Then repeat the previous formulae I = sqrt(P/R) with the new resistance.

I = sqrt 1/10500
I = sqrt 9.523
I = 9.76 Ma

Thank you for your help. I often get mixed up with the units and make very basic mistakes, something i need to work on thanks..
 
licklecee said:
oh okay,

1.so sqrt 0.00001 = 3.16

2. Would it be 3.16 amps then ?

3. Okay so is it correct to say that to calculate maximum current, i need to calculate the maximum resistance, using the tolerance of in this case 5 percent,. Then repeat the previous formulae I = sqrt(P/R) with the new resistance.

I = sqrt 1/10500
I = sqrt 9.523
I = 9.76 Ma

Thank you for your help. I often get mixed up with the units and make very basic mistakes, something i need to work on thanks..

1. recalculate sqrt(1/10000) and watch your decimal places.

2. No.
 
Also to calculate the maximum voltage, would it be correct to say ...

V=IR
V = 9.76*9500 ( 9500 because that is the lowest resistance point )
V = 0.00976*9500
V = 92.72 volts

Thank you
 
Thread 'Have I solved this structural engineering equation correctly?'
Hi all, I have a structural engineering book from 1979. I am trying to follow it as best as I can. I have come to a formula that calculates the rotations in radians at the rigid joint that requires an iterative procedure. This equation comes in the form of: $$ x_i = \frac {Q_ih_i + Q_{i+1}h_{i+1}}{4K} + \frac {C}{K}x_{i-1} + \frac {C}{K}x_{i+1} $$ Where: ## Q ## is the horizontal storey shear ## h ## is the storey height ## K = (6G_i + C_i + C_{i+1}) ## ## G = \frac {I_g}{h} ## ## C...

Similar threads

Replies
1
Views
5K
Replies
5
Views
2K
Replies
11
Views
3K
Replies
9
Views
2K
Replies
6
Views
2K
Replies
1
Views
1K
Replies
2
Views
12K
Back
Top