Two resistors R1 and R2 are connected parallel across points X and Y. The current
entering X and leaving Y are equal and is I. Derive the expression for the current through
R1 and R2 in terms of I.
Answers
Answer:
Solution
If R_{1} and R_{2} connect in series
. R eq =R 1 +R 2
So, R s =R 1 +R 2 ...(1)(R s =R eq in series) If R_{1} and R_{2} connect in parallel
therefore 1 R eq = 1 R 1 + 1 R 2
Rightarrow 1 R eq = R 1 +R 2 R 1 R 2 Rightarrow R eq = R 1 +R 2 R 1 R 2
So, Rightarrow R p = R 1 +R 2 R 1 R 2 ...(2)(R p =R eq
parallel)
From equation (1) and (2)
R_{p} = (R_{1}*R_{2})/R_{3} R_{s} > R_{p}
Now,
I= V R Rightarrow I propto 1 R
Since R$ > Rup, Is = current in series Ip = current in parallel
Answer:
Two resistors, R
1
andR
2
are connected in parallel.
We know,
R
e
1
=
R
1
1
+
R
2
1
Then R
e
=
R
1
+R
2
R
1
R
2
=
50+100
50×100
=
150
5000
=
3
100
Now, Parallel Connection error:
=
(R
1
+R
2
)
2
R
1
2
(dB)+R
2
2
(dA)
=
150
2
50
2
(3)+100
2
(2)
=
150×150
7500+20000
=
9
11
Relative Error =
(100/3)
(11/9)
=0.03666