A 60 W electric light bulb is connected to a 220 V supply. Determine (a) the current flowing in the bulb, and (b) the resistance of the bulb ?

 Solution: To determine the current flowing in the bulb and the resistance of the bulb, we can use Ohm's law and the formula for power in terms of voltage and resistance.

(a) Current flowing in the bulb:

We can use Ohm's law to find the current ($𝐼$) flowing in the bulb, which states:

$𝑃=𝑉𝐼$

Where:

• $𝑃$ is the power (60 W),
• $𝑉$ is the voltage (220 V),
• $𝐼$ is the current.

Rearranging the formula to solve for $𝐼$:

$𝐼=\frac{𝑃}{𝑉}$

Plugging in the given values:

$𝐼=\frac{60\text{W}}{220\text{V}}$

$𝐼\approx 0.273\text{A}$

So, the current flowing in the bulb is approximately $0.273\text{A}$.

(b) Resistance of the bulb:

We can use the relationship between power ($𝑃$), voltage ($𝑉$), and resistance ($𝑅$), which is:

$𝑃=\frac{{𝑉}^2}{𝑅}$

Rearranging the formula to solve for $𝑅$:

$𝑅=\frac{{𝑉}^2}{𝑃}$

Plugging in the given values:

$𝑅=\frac{(220\text{V}{)}^2}{60\text{W}}$

$𝑅=\frac{48400{\text{V}}^2}{60\text{W}}$

$𝑅\approx 806.67\mathrm{\Omega }$

So, the resistance of the bulb is approximately $806.67\mathrm{\Omega }$.