In the figure R₁ = R₂=3.60, R3=4.80, R4 = 5.60, R5 = 3.402, and R6 = R7 = Rg = 10.30, and the ideal batteries have emfs 81 = 27 V and 82= 10.9 V. What are the (a) magnitude and (b) direction (up or down) of current in and the (c) magnitude and (d) direction of current i2? What is the energy transfer rate in (e) battery 1 and (f) battery 2? Is energy being supplied or absorbed in (g) battery 1 and (h) battery 2? R₁ Re www R₂ R₂ Ro www Rg 8₁ 4 R₁ 88%87 04 R₂

icon
Related questions
Question
=
In the figure R₁ = R₂ = 3.60, R3 = 4.8 Q, R4 = 5.6 0, R5 = 3.4 02, and R6 R7 = Rg = 10.30, and the ideal batteries have emfs 8₁ = 27 V and
82 = 10.9 V. What are the (a) magnitude and (b) direction (up or down) of current in and the (c) magnitude and (d) direction of current
12? What is the energy transfer rate in (e) battery 1 and (f) battery 2? Is energy being supplied or absorbed in (g) battery 1 and (h)
battery 2?
(a) Number
(b)
(c) Number
(d)
(e) Number
(f) Number
IN
>
jed
T
www
www
R₁
Rs
Units
Units
Units
Units
www
R₂
2
www
R7
Ro
www
www w
R₁
Rg
2
&17
83
R
www
Transcribed Image Text:= In the figure R₁ = R₂ = 3.60, R3 = 4.8 Q, R4 = 5.6 0, R5 = 3.4 02, and R6 R7 = Rg = 10.30, and the ideal batteries have emfs 8₁ = 27 V and 82 = 10.9 V. What are the (a) magnitude and (b) direction (up or down) of current in and the (c) magnitude and (d) direction of current 12? What is the energy transfer rate in (e) battery 1 and (f) battery 2? Is energy being supplied or absorbed in (g) battery 1 and (h) battery 2? (a) Number (b) (c) Number (d) (e) Number (f) Number IN > jed T www www R₁ Rs Units Units Units Units www R₂ 2 www R7 Ro www www w R₁ Rg 2 &17 83 R www
Expert Solution
Step 1

Since we answer up to three subparts, we will answer the first three. Please resubmit the question and specify the other subparts you would like to get answered. 

The resistors R7 and R8 are connected in parallel, so the equivalent resistance of these two resistors are 

R78=R8R7R8+R7R78=10.3 Ω×10.3 Ω10.3 Ω+10.3 ΩR78=5.15 Ω

Now the resistors R78, R1 and R2 are in series, the equivalent resistance is 

R7812=R78+R1+R2R7812=5.15 Ω+3.6 Ω+3.6 ΩR7812=12.35 Ω

Now R7812 and R6 is parallel, the equivalent resistance is 

R78126=R7812R6R7812+R6R78126=12.35 Ω×10.3 Ω12.35 Ω+10.3 ΩR78126=5.62 Ω

Now the circuit becomes 

Advanced Physics homework question answer, step 1, image 1

Here I1 and I2 are the currents in loop

 

steps

Step by step

Solved in 2 steps with 1 images

Blurred answer