Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

See Figure 1 - http://imgur.com/XEEAixN

ID: 2140454 • Letter: S

Question


Part A
Find the equivalent resistance of the entire network. Set R1 = 18? , R2 = 11? , R3 = 27? , R4 = 65? .
Req = 42 Ohms

Part B
Find the current through each resistor.
I1,I2,I3,I4 = ?

Part C
Find the potential difference across each resistor.
V1,V2,V3,V4 = ?

Express all answers to two significant figures.

Any help is appreciated See Figure 1 - http://imgur.com/XEEAixN
Part A
Find the equivalent resistance of the entire network. Set R1 = 18? , R2 = 11? , R3 = 27? , R4 = 65? .
Req = 42 Ohms

Part B
Find the current through each resistor.
I1,I2,I3,I4 = ?

Part C
Find the potential difference across each resistor.
V1,V2,V3,V4 = ?

Express all answers to two significant figures.

Any help is appreciated Find the equivalent resistance of the entire network. Set R1 = 18? , R2 = 11? , R3 = 27? , R4 = 65? . Req = 42 Ohms Find the current through each resistor. I1,I2,I3,I4 = ? Find the potential difference across each resistor. V1,V2,V3,V4 = ? Express all answers to two significant figures. Any help is appreciated

Explanation / Answer

first combine R2 and R3
R23 = 11 + 27 = 38

combine R23 and R4

1/R = 1/38 + 1/65

R234 = 23.98

Rtotal = 18 + 23.98 = 42

B)

I total = 10/42=
0.238

things in series have same current so I1 = 0.24
and thus V1 = 0.238*18=4.3

V234 = 0.238*23.98=5.7 V

so V4 = 5.7 and thus I4 = 5.707/65=0.088 A

then I23 = 5.707/38=0.15 A

so I2 =I3 = 0.150 A

V2 = 0.15 *11 = 1.7 V

V3 = 0.15*27= 4.1 V

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote