physics 102 circuit. pts will be given to best answer Two resistors R1 = 60 omeg

Question

physics 102 circuit. pts will be given to best answer

Two resistors R1 = 60 omega and R2 = 60 omega are connected in parallel. This parallel arrangement is connected in series with resistor R3 = 3 omega. The combination is then placed across a 120 V potential difference. a) (3 points) What is the equivalent resistance that could replace the three original resistors? b) (2 points) What is the current in the circuit? c) (3 points) What is the voltage across the 30 omega resistor? d) (3 points)What is the voltage across the parallel portion of the circuit? e) (3 points) What is the current in each branch of the parallel portion of the circuit? f) (3 points) What power is released on each of the resistors?

Explanation / Answer

A)

R1,R2 are connected in parllel

resultant resistance will be=R1*R2/R1+R2

=60*60/120

=30ohms

30ohms is connected in series with R3

then equivalent resistance=30+R3

=30+30

=60ohms

B)

I=V/R

=120/60

=2A

C)

voltage across 30 ohm resistor

V=IR

=30*2

=60V

D)

apply kvl law

120=60+voltage across parllel portion

voltage across parllel portion

=60V

E)

current in each branch

I=60/60

=1A

F)

P=I2*R

using above concept

P1=1*60

=60W

P2=1*60

=60W

P3=4*30

=120W

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.