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

1) If the potential difference across the filament is 120 V, what is the strength of the electric field inside the filament?

2) Suppose the length of the bulb’s filament was doubled without changing its diameter or the potential difference across it. What would the electric field strength be in this case?

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?

4) What is the resistance of the filament following the doubling of its length?

**Step-by-step explanation**

### 1) If the potential difference across the filament is 120 V, what is the strength of the electric field inside the filament?

Given that, length of filament, l = 60 cm = 0.6m

Potential difference across the filament is, V = 120 V

We know, the electric field , E = V/l

=120/0.6

=** 200 V/m**

### 2) Suppose the length of the bulb’s filament was doubled without changing its diameter or the potential difference across it. What would the electric field strength be in this case?

If the length is doubled then the new length , L = 2*0.6 = 1.2 m

The electric field is,

E1 = V/L

= 120/1.2 V/m

**= 100 V/m**

Or, the electric field is,

E1=E/2= 200/2 V/m

**= 100 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?

We know, Power P = 60 W and Potential Difference, V = 120 V

The current flowing through the filament,

I= P/V

=60/120

=0.5A

Since the current is directly proportional to the electric field, the current will become half of the initial current. As the electric field became half after doubling the filament length.

So, the new current is,

I1 = I/2

= 0.5/2 A

**=0.25 A**

Lets Verify,

As Potential Difference was kept unchanged and it is V = 120 V and doubling length doubles the filament resistance that is R1 = 2*240 Ohms = 480 Ohms.

So, the new current, I1 = V/R1 = 120/480 =** 0.25 A **

### 4) What is the resistance of the filament following the doubling of its length?

Since the resistance is directly proportional to the length.

The new resistance is R1=2*240 Ohms

**=480 ohms**

## Glossary

**Electric Field**

Electric field is the electric force per unit charge. The symbol the electric field is **E**. Unit is volts per meter (V/m) .

More about Electric Field.

**Why doubling the length of the filament doubles the resistance? If filament diameter is unchanged.**

The resistance of a given object depends primarily on two factors: What material it is made of, and its shape. For a given material, the resistance is inversely proportional to the cross-sectional area; for example, a thick copper wire has lower resistance than an otherwise-identical thin copper wire. Also, for a given material, the resistance is proportional to the length; for example, a long copper wire has higher resistance than an otherwise-identical short copper wire.

where, l is the length of the conductor, measured in metres (m), A is the cross-sectional area of the conductor measured in square metres (m^{2}), and ρ (rho) is the electrical resistivity.

**Ohm’s law**

The current I through the material is proportional to the voltage V applied across it:

So, we have solved the problem of **The length of a 60 W, 240 ω light bulb filament is 60 cm.** Hope you’ve understood the explanation of the problem.