.GU- 4.0V 2. - -= = i E=Lav R₁=10_n E=3.0V R₂ = 10-12 w R2=540 M Rp -10.0_1 انتها Ru= 68.0_n N Ro=68.01

University Physics Volume 3
17th Edition
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:William Moebs, Jeff Sanny
Chapter9: Condensed Matter Physics
Section: Chapter Questions
Problem 80P: What is the critical magnetic field for lead at T = 2.8 K?
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The voltage for the first battery is 4.0 V. The voltage for the second one is 3.0 V indicated in the circuit diagram.

The Resistance 1 is 10 ohm

Resistance 2 is 51 ohm

Resistance 3 is 10 ohm

Resistance 4 is 68 ohm

resistance 5 is 68 ohm

resistance 6 is 10 ohm

Can you help calculate the total current running through the circuit diagram? Thanks in advance. Please explain it step by step without short cut. I need to understand this please.

 
.GU-
4.0V
2.
-
-=
= i
E=Lav
R₁=10_n
E=3.0V
R₂ = 10-12
w
R2=540
M
Rp -10.0_1
انتها
Ru= 68.0_n
N
Ro=68.01
Transcribed Image Text:.GU- 4.0V 2. - -= = i E=Lav R₁=10_n E=3.0V R₂ = 10-12 w R2=540 M Rp -10.0_1 انتها Ru= 68.0_n N Ro=68.01
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