Question

an electric bulb is rated at 60W, 240V. calculate its resistance. if the voltage drops to 192 v, calculate the power consumed and the current drawn by the bulb.​

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Power, (P) = 60 W

Potential difference, (V) = 240 V

We know that,

P = V²/R

⇒ 60 = (240)²/R

⇒ R = 57600/60

⇒ R = 960 Ω

Here, the voltage of the bulb falls to 192 V.

Resistance = 960 Ω

We know that,

V = IR

⇒ 192 = I/960

⇒ I = 192/960

⇒ I = 0.2 A

Hence, the current drawn is 0.2 A.

Power = Current × Voltage

⇒ P = 0.2 × 192

⇒ P = 38.4 W

Hence, the power consumed is 38.4 W.

Answer 2

Given

➝ Power of bulb = 60 W

➝ Potential difference = 240 V

To find

➝ Resistance of bulb

➝ Power consumed and current drawn by bulb when voltage drops to 192 v

Solution

At first to find resistance, we will use this formula :

➝ Power(P) = Voltage (V)²/Resistance (R)

➝ 60 = (240)²/R

➝ 60R = 57600

➝ R = 57600/60

➝ R = 960Ω

∴ Resistance of bulb = 960Ω

___________________________

Now we have,

➝ Voltage (V) = 192v

➝ Resistance (R) = 960Ω

Now using same formula :-

➝ P = V²/R

➝ P = (192)²/960

➝ P = 36864/960

➝ P = 38.4 Watts.

∴ Power consumed by bulb = 38.4 Watts.

___________________________

Now to find current drawn by bulb we will use Ohm's law :

➝ Potential difference (V) = Current (I) * Resistance (R)

➝ 192 = I × 960

➝ I = 192/960

➝ I = 0.2 A

∴ Current drawn by bulb = 0.2 Amperes.