Question

A lamp wich is connected to 230v draws a current of 0.261A calculate the power and hot resistence of the lamp?​

Answer 1

Answer:

• Power of Lamp is 60 Watt.
• Resistance of lamp is 881.22 Ohms

Explanation:

Given:-

• Potential Difference ,P = 230 v
• Current ,I = 0.261 A

To Find:-

• Power of Lamp ,P
• Resistance ,R

Solution:-

Firstly we calculate the Resistance of the Lamp.

Using Ohm's Law

• V = IR

where,

• V denote Potential Difference
• I denote Current
• R denote Resistance

Substitute the value we get

→ 230 = 0.261 × R

→ R = 230/0.261

→ R = 230000/261

→ R = 881. 22Ω

• Hence, the Resistance of the Lamp is 881.22 ohms.

Now, calculating the Power

• P = V²/R

Substitute the value we get

→ P = 230²/881.22

→ P = 230×230/881.22

→ P = 52900/881.22

→ P = 60.03 ≈ 60 W

• Hence, the Power of the Lamp is 60 Watt.

Answer 2

QuestioN :-

⠀⠀

• A lamp which is connected to 230v draws a current of 0.261A. Calculate the power and hot resistance of the lamp.

⠀

Given :-

A lamp is connected to 230v draws a current of 0.261A

So,

• the Potential Difference (P) = 230v

• Current (i) = 0.261 A

To FinD :-

• The power of lamp

• Hot resistance (r)

⠀

AnsweR :-

First , Let's find out the Resistance of the lamp :-

we know that :-

• V = ir ( Ohm's law )

Where ,

$\star \sf \: v = donate \: potential \: difference \\ \\ \star \sf \: i = donate \: current \\ \\ \star \sf \: r = donate \: resestance$

Putting the values we have into the ohm's law :-

$\implies \sf \: v = ir \\ \\ \implies \sf \: 230 = 0.261 \times r \\ \\ \implies \sf \: r = \frac{230}{0.261} \\ \\ \implies \sf r = 881.22 \: ohm \\ \\ \sf so \: the \: resistence \: of \: the \: lamp \: is \: 881.22 \: ohms$

Now , to FinD the the Power :-

Using the identity -

• $\sf \: p = \frac{ {v}^{2} }{r}$

$\implies \sf \: p = \frac{ {(230)}^{2} }{881.22} \\ \\ \implies \sf \: p = \frac{52900}{881.22} \\ \\ \implies \boxed{\sf \: p = 60.03 \:w}$

Hence , the Resistance = 881.22 ohm's and power = 60.03 w.

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Hope it will help you :)