Question

anyone can help me in physics class 12th??A lamp consumes 500 J of electrical energy in 20 seconds. What is the power of the lamp?

Also find the electrical energy consumed in units if the lamp operates for 2 hours daily for 15 days.​

Answer 1

Answer:

• 25 Watt is the Power of Lamp.
• 1.38 × 10⁻⁴ kWh is the required answer

Explanation:

Given:-

• Energy Consumes ,E = 500 J

• Time taken ,t = 20 s

To Find:-

• Power of lamp ,P.

Solution:-

Energy Consumes by a body is defined as the product of power of appliances & time taken.

• E = P× t

Substitute the value we are

→ 500 = P × 20

→ P = 500/20

→ P = 50/2

→ P = 25 watt.

• Hence, the power of the Lamp is 25 Watts.

Now, we have to calculate the energy consumed by lamp which operates for 2 hour daily for 15 days.

• E = P× t

Substitute the value we get

→ E = 25 × (2 × 15)

→ E = 25 × 30

→ E = 500 J

As we know that 1 Unit or kWh = 3.6×10⁶ J

Therefore, 500 J = 500/3.6×10⁶

→ 500×10⁻⁶ /3.6 unit

→ 138 × 10⁻⁶ unit

→ 1.38 × 10⁻⁴ Unit

• Hence, the energy consumed by the Lamp in Unit is 1.38 ×10⁻⁴ unit or kWh.

Answer 2

Explanation:

$\frak Given = \begin{cases} &\sf{Energy\ consumes,\ E\ = \bf{500\ J}} \\ &\sf{Time\ taken,\ t\ = \bf{20sec}} \end{cases}$

To find:- We have to find the power of the lamp, P ?

___________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf\pink{\underline{E\ =\ P\ ×\ t.}}$

Here:-

● E is for energy consumed.

● P is for power.

● t is for time taken.

___________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values:-}}}$

$\sf : \implies {500\ =\ P\ ×\ 20} \\ \\ \sf : \implies {P\ =\ \dfrac{500}{20}} \\ \\ \sf : \implies {P\ =\ \dfrac{50}{2}} \\ \\ \sf : \implies {\purple{\underline{P\ =\ 25watt.}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ power\ of\ the\ lamp\ is\ 25\ Watts.}}$

____________________

$\frak{\underline{\underline{\dag Now,\ we\ will\ calculate\ the\ energy\ consumed\ by\ the\ lamp:-}}}$

$\sf{\underline{\underline{By\ substituting\ the\ values:-}}}$

$\sf\pink{\underline{E\ =\ P\ ×\ t.}}$

$\sf : \implies {E\ =\ 25\ ×\ (2\ ×\ 15)} \\ \\ \sf : \implies {E\ =\ 25\ ×\ 30} \\ \\ \sf : \implies {E\ =\ 500\ J}$

___________________

$\frak{\underline{\underline{\dag We\ know\ that:-}}}$

☯️ 1 unit or kWh = 3.6 × 10⁶ J.

$\sf \therefore {500\ J\ =\ \dfrac{500}{3.6×10⁶}}$

$\sf : \implies {\dfrac{500×10^‐⁶}{3.6\ unit}} \\ \\ \sf : \implies {138\ ×\ 10^‐⁶\ unit} \\ \\ \sf : \implies {\purple{\underline{1.38\ ×\ 10^‐⁴\ unit.}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ energy\ consumed\ by\ the\ lamp\ in\ units\ is\ 1.38\ ×\ 10^‐⁴.}}$