The length of a 60 W, 240 ohms light bulb filament is 60 cm. 1) If the potential

Question

The length of a 60 W, 240 ohms light bulb filament is 60 cm.

1) If the potential difference across the filament is 120 W, what is the strength of the electric field inside the filament? E = V/m
2) Suppose the length of the bulb's filament were doubled without changing its diameter or the potential difference across it. What would the electric field strength be in this case? E = V/m
3) Remembering that the current in the filament is proportional to the electric field, what is the current in the filament following the doubling of its length? I = A
4) What is the resistance of the filament following the doubling of its length? R = ohms

Explanation / Answer

Given that length l=0.6m

potential difference is V=120V

hence the electric field is E=V/l

=120/0.6= 200 V/m

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

if the length is doubled then the electric field is

E1=E/2=100 V/m

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

intial current is I=P/V

=60/120

=0.5A

since the current is directly preportinal to electric field current also reduces by a factor 2

that is the new current is I2=I/2=0.25A

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

since the resistance is directly preportional to the length

the new resistance is R2=2(240)=480 ohm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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