Physics, asked by muskanchaurasia19, 7 months ago

निम्न परिपथ में बहने वाली विद्युत धारा । की गणना कीजिए।
12 वोल्ट
12
ww
कुंजी
11
12
42
H
I 22 I II 12
(उ०प्र०20​

Answers

Answered by anubhabkumar2020
0

Answer:

Answer:

Answer:

Answer: L=2m,

Answer: L=2m,d=3mm,A=

Answer: L=2m,d=3mm,A= 4

Answer: L=2m,d=3mm,A= 49π

Answer: L=2m,d=3mm,A= 49π

Answer: L=2m,d=3mm,A= 49π ×10

Answer: L=2m,d=3mm,A= 49π ×10 −6

Answer: L=2m,d=3mm,A= 49π ×10 −6 m

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .

Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .

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