Question

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference.

A.What is the resistance of a 100

W bulb?

B.How much current does the 100

W bulb draw in normal use?

C.What is the resistance of a 60

W bulb?

D.How much current does the 60

W bulb draw in normal use?

Answer 1

Concepts and reason

The main concept used is the relation between the power and voltage.

Use the formula of power to calculate resistance and current.

Fundamentals

Formula to calculate power is given below.

$P = VI$

And

$P = \frac{{{V^2}}}{R}$

Here, $P$ is power of source, $V$ is voltage, $I$ is current and $R$ is resistance.

(A)

Calculate the resistance of 100W bulb.

Rearrange the formula given above for resistance.

$R = \frac{{{V^2}}}{P}$

Substitute, $120{\rm{ V}}$ for $V$ and $100{\rm{ W}}$ for $P$.

$\begin{array}{c}\\R = \frac{{{{\left( {120{\rm{ V}}} \right)}^2}}}{{100{\rm{ W}}}}\\\\ = 144{\rm{ }}\Omega \\\end{array}$

(B)

Calculate the Current drawn by the 100 W bulb in normal use.

Rearrange the formula of power given above, for current.

$I = \frac{P}{V}$

Substitute, $100{\rm{ W}}$ for $P$ and $120{\rm{ V}}$ for $V$.

$\begin{array}{c}\\I = \frac{{100{\rm{ W}}}}{{120{\rm{ V}}}}\\\\ = 0.83{\rm{ A}}\\\end{array}$

(C)

Calculate the resistance of 60 W bulb.

Rearrange the formula given above for resistance.

$R = \frac{{{V^2}}}{P}$

Substitute, $120{\rm{ V}}$ for $V$ and $60{\rm{ W}}$ for $P$.

$\begin{array}{c}\\R = \frac{{{{\left( {120{\rm{ V}}} \right)}^2}}}{{60{\rm{ W}}}}\\\\ = 240{\rm{ }}\Omega \\\end{array}$

(D)

Calculate the current drawn by 60 W bulb.

Rearrange the formula of power given above, for current.

$I = \frac{P}{V}$

Substitute, $60{\rm{ W}}$ for $P$ and $120{\rm{ V}}$ for $V$.

$\begin{array}{c}\\I = \frac{{60{\rm{ W}}}}{{120{\rm{ V}}}}\\\\ = 0.50{\rm{ A}}\\\end{array}$

Ans: Part A

Resistance of bulb is $144{\rm{ }}\Omega$.

Part B

The bulb will draw $0.83{\rm{ A}}$ current.

Part C

Resistance of bulb is $240{\rm{ }}\Omega$.

Part D

The bulb will draw $0.50{\rm{ A}}$ current.