saifullah99319
Member
LOCUS,
Sorry, a had a problem of my opinion.
steps:
1.First measure power supply, let it is (E) volt
and current rating is (I) ampere > 0.75amp.
2. Let take power supply source resistance =Rs ohm.
Fuse impedence =Z1 ohm.
Load impedence =Z2 ohm.
You choose 600v-0.75amp fuse.ok.
Significance of 0.75amp of fuse:
1.let, 0.75amp =E/(Rs+Z1+Z2).............(A)
Rs=0 ohm.(assume)
Z1=constant(by ohm meter)
z2 = (E/0.75)-Z1 ohm.
So this Z2 ohm is minimum value for maximum of 0.75ampere.
Under calculation, if you insert Z2 ohm then I1=0.75 ampere
will pass through the circuits.
conclusion: So under calculation of E,Z1
you can insert Z2 ohm of device.(for max of 0.75amp)
After knowing Z2 value you can insert higher
than Z2 value then current will less than 0.75amp.
Significance of 600v of fuse:
1. But if you insert a device whose impedence is less than
Z2 ohm then by equation (A), (greater than 0.75 ampere)=I2 will
circulate through the circuit.
2. then I2 squared Z1 is relatively larger than I1 squared Z1.
3. So in this case fuse will be heated. and tempereture
will rise.
4.Let we assume fuse element is positive tempereture coefficient.
so if temp rises then resistance rises by Rt=R0(1+alpha*t).
where,
Rt = final resistance.
R0 = initial resistance.
alpha = tempereture coefficient of fuse element.
t= tempereture difference.
since, rising current is constant i.e.,greater than 0.75amp
voltage accross the fuse will be V=I2*Z1.
5. If voltage accross the fuse is (V=I2*z1) greater than 600V
then fuse melts.
conclusion:
So you must insert of Z2 impedence for 0.75ampere.(under
calculation)
[ nb. It is also true that if fuse current rating is greater than
power supply rating then powersupply may dammage.
But no effect on fuse.]
Sorry, a had a problem of my opinion.
steps:
1.First measure power supply, let it is (E) volt
and current rating is (I) ampere > 0.75amp.
2. Let take power supply source resistance =Rs ohm.
Fuse impedence =Z1 ohm.
Load impedence =Z2 ohm.
You choose 600v-0.75amp fuse.ok.
Significance of 0.75amp of fuse:
1.let, 0.75amp =E/(Rs+Z1+Z2).............(A)
Rs=0 ohm.(assume)
Z1=constant(by ohm meter)
z2 = (E/0.75)-Z1 ohm.
So this Z2 ohm is minimum value for maximum of 0.75ampere.
Under calculation, if you insert Z2 ohm then I1=0.75 ampere
will pass through the circuits.
conclusion: So under calculation of E,Z1
you can insert Z2 ohm of device.(for max of 0.75amp)
After knowing Z2 value you can insert higher
than Z2 value then current will less than 0.75amp.
Significance of 600v of fuse:
1. But if you insert a device whose impedence is less than
Z2 ohm then by equation (A), (greater than 0.75 ampere)=I2 will
circulate through the circuit.
2. then I2 squared Z1 is relatively larger than I1 squared Z1.
3. So in this case fuse will be heated. and tempereture
will rise.
4.Let we assume fuse element is positive tempereture coefficient.
so if temp rises then resistance rises by Rt=R0(1+alpha*t).
where,
Rt = final resistance.
R0 = initial resistance.
alpha = tempereture coefficient of fuse element.
t= tempereture difference.
since, rising current is constant i.e.,greater than 0.75amp
voltage accross the fuse will be V=I2*Z1.
5. If voltage accross the fuse is (V=I2*z1) greater than 600V
then fuse melts.
conclusion:
So you must insert of Z2 impedence for 0.75ampere.(under
calculation)
[ nb. It is also true that if fuse current rating is greater than
power supply rating then powersupply may dammage.
But no effect on fuse.]