class 11 units and measurements...
Attachments:
Answers
Answered by
4
CASE 1: Series Combination
For series combination equivalent resistance R = R1 + R2
= R = 100 ± 3 + 200 ± 4 = R = 300 ± 7 ohm
As for addition or substraction the errors are simply added together.
CASE 2: Parallel combination
For parallel combination equivalent resistance is given by
1/R = 1/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2)
= R = (R1x R2)/(R1+ R2)
=Thus, R = (100 x 200)/(100 + 200)
= R = 20000/300 = 200/3
= R = 66.67 ohm
Now to calculate error we use
∆R/R2 = ∆R1/R12+ ∆R2/R22
Putting values we get
∆R/(66.67)2 = 3/(100)2 + 4/(200)2
= ∆R/(66.67)2= 3/10000 + 4/40000
= ∆R/(66.67)2 = 3/10000 + 1/10000
= ∆R/(66.67)2 = 4/10000
= ∆R = (4/10000) x (66.67)2
= ∆R = 4/10000 x 4444.889
= ∆R = 17779.56/10000
= ∆R = 1.7779 = ∆R = 1.78 Ohm
Thus, total equivalent resistance in parallel combination
= R ± ∆R
= 66.67 ± 1.78 Ohm
For series combination equivalent resistance R = R1 + R2
= R = 100 ± 3 + 200 ± 4 = R = 300 ± 7 ohm
As for addition or substraction the errors are simply added together.
CASE 2: Parallel combination
For parallel combination equivalent resistance is given by
1/R = 1/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2)
= R = (R1x R2)/(R1+ R2)
=Thus, R = (100 x 200)/(100 + 200)
= R = 20000/300 = 200/3
= R = 66.67 ohm
Now to calculate error we use
∆R/R2 = ∆R1/R12+ ∆R2/R22
Putting values we get
∆R/(66.67)2 = 3/(100)2 + 4/(200)2
= ∆R/(66.67)2= 3/10000 + 4/40000
= ∆R/(66.67)2 = 3/10000 + 1/10000
= ∆R/(66.67)2 = 4/10000
= ∆R = (4/10000) x (66.67)2
= ∆R = 4/10000 x 4444.889
= ∆R = 17779.56/10000
= ∆R = 1.7779 = ∆R = 1.78 Ohm
Thus, total equivalent resistance in parallel combination
= R ± ∆R
= 66.67 ± 1.78 Ohm
NidhraNair:
Anyways thanks ☺
I not copy anything
your welcome
there are two methods
one for board and other for competition
board method doesn't work for JEE and vica versa
in the anger your English is fine. but in commemts y unpolished ????
yep it's short cut ☺
whose?
jee
Answered by
5
CASE 1: Series Combination
For series combination equivalent resistance R = R1 + R2
= R = 100 ± 3 + 200 ± 4 = R = 300 ± 7 ohm
As for addition or substraction the errors are simply added together.
CASE 2: Parallel combination
For parallel combination equivalent resistance is given by
1/R = 1/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2)
= R = (R1x R2)/(R1+ R2)
=Thus, R = (100 x 200)/(100 + 200)
= R = 20000/300 = 200/3
= R = 66.67 ohm
Now to calculate error we use
∆R/R2 = ∆R1/R12+ ∆R2/R22
Putting values we get
∆R/(66.67)2 = 3/(100)2 + 4/(200)2
= ∆R/(66.67)2= 3/10000 + 4/40000
= ∆R/(66.67)2 = 3/10000 + 1/10000
= ∆R/(66.67)2 = 4/10000
= ∆R = (4/10000) x (66.67)2
= ∆R = 4/10000 x 4444.889
= ∆R = 17779.56/10000
= ∆R = 1.7779 = ∆R = 1.78 Ohm
Thus, total equivalent resistance in parallel combination
= R ± ∆R
= 66.67 ± 1.78 Ohm
For series combination equivalent resistance R = R1 + R2
= R = 100 ± 3 + 200 ± 4 = R = 300 ± 7 ohm
As for addition or substraction the errors are simply added together.
CASE 2: Parallel combination
For parallel combination equivalent resistance is given by
1/R = 1/R1 + 1/R2 = 1/R = (R1 + R2)/(R1x R2)
= R = (R1x R2)/(R1+ R2)
=Thus, R = (100 x 200)/(100 + 200)
= R = 20000/300 = 200/3
= R = 66.67 ohm
Now to calculate error we use
∆R/R2 = ∆R1/R12+ ∆R2/R22
Putting values we get
∆R/(66.67)2 = 3/(100)2 + 4/(200)2
= ∆R/(66.67)2= 3/10000 + 4/40000
= ∆R/(66.67)2 = 3/10000 + 1/10000
= ∆R/(66.67)2 = 4/10000
= ∆R = (4/10000) x (66.67)2
= ∆R = 4/10000 x 4444.889
= ∆R = 17779.56/10000
= ∆R = 1.7779 = ∆R = 1.78 Ohm
Thus, total equivalent resistance in parallel combination
= R ± ∆R
= 66.67 ± 1.78 Ohm
Similar questions