Problem 21 (12 points) Six resistor are connected as shown to a power supply wit

Question

Problem 21 (12 points) Six resistor are connected as shown to a power supply with emf & and zero internal resistance. The resistances are R- (a) (3 pts) In terms of E and R, determine 14, the current through resistor Re. (b) (2 pts) In terms of , determine AVa, the potential R s difference across resistor R (c) (3 pts) Consider the points marked A and B in the circuit. R6 6) Which point, A or B, is at the higher potential? (i) In terms of E, determine V,, the magnitude of the potential difference between points A and B. (d) (4 pts) Assume each of the six resistors is a light bulb with its resistance constant at the values given above. () Would the brightness of bulb I be greater than lessthan, or equal to the brightness of bulb 6? Explain your reasoning (i) Would the brightness of the bulb 4 be greater than, less than, or equal to the brightness of bulb 52 Explain your reasoning.

Explanation / Answer

R2 , R3 are in series R23 = R2 + R3 = 2R+ 6R = 8R

R4 , R5 are in series R45 = R4 + R5 = 6R + 2R = 8R

R23 and R45 are in paralle R2345 = (R23*R45)/(R23 + R45)

R2345 = 4R

R1 , R2345 , R6 are in series

Rtot = R1 + R2345 + R6

Rtot = 6 R

current through R1 , I1 = E/Rtot = E/(6R)

potential across R45, V45 = I1*R2345 = E*(2/3)

current in R4 , I4 = V45/R45 = E/12R

============================

(b)

V4 = I4*R4 = E/2

====================

(C)

(i)

A is high potential

---------------

(ii)

vA + I2*R2 - I4*R4 = VB

VA + (E/12R)*2R - (E/12R)*6R = vB

VA + E/6 - E/2 = VB

VA - VB = E/3

==========================

d)

equal

power = I^2*R

current and resistance of 1 and 6 are same

(ii)

same current flows through R4 , R5

R4 > R5

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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