The four light bulbs in the circuit shown all have the same resistance. (a) Rank

Question

The four light bulbs in the circuit shown all have the same resistance. (a) Rank the brightness of the bulbs in order of brightness, separated by greater than or equal signs (like 4 > 3 = 1 = 2). Explain your reasoning. (b) If the resistance of each bulb is 5 Ohm and the battery provides a voltage of 10 V, what is the total current passing through the battery? (c) How many electrons leave the battery in one minute? (d) The resistor in bulb 1 is made of a cylindrical piece of material with a length of 5 cm and a diameter of 0.1 mm. What is the resistivity of the material?

Explanation / Answer

(a) The resistance of two light bulbs in parallel is smaller than that of two light bulbs in series. Thus, the current through the battery is greater for bulb1 than for the bulbs connected in series. Since the power dissipated is the product of current and potential difference, it follows that more power is dissipated in bulb 1.

Hence 1>2=3=4

(b) Bulbs 2 to 4 are connected series Rs = R2+R3+R4 where each bulb is 5, Rs = 15 which is in parallel connection with R1=5 resistance of bulb1

R = (1/Rs)+(1/R1)=0.27

voltage provided by battery V=10 V

Then total current through the battery is I=V/R = 10/0.27 = 3.70 A

(c) V = I R (potential = current x resistance)
    Q = I t (charge = current x time)
    Ne = Q / e = I t / e = (V / n) (t / e)
    Ne = (10 V)(60 s) / [(0.27 ohms) (1.6 x 10-19C)] = 1.38 x 1022 electrons per minute

(d) R= L/A , where is the resistivity

A is cross sectional area of the wire = r2 = 3.14x(5x10-5)2 = 7.85x10-9 m2

Length of the wire, L= 5x10-2 m

= RA/L = (5x7.85x10-9 )/5x10-2= 7.85x10-7 .m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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