Question

b) Write two points of difference between electric energy and electric power. Two incandescent light bulbs draw 60 W and 40 W individually when connected across 240 V DC supply, what are their resistances ?​

Answer 1

$\huge\bold{QUESTION}$

❥Write two points of difference between electrical energy and electric power.

❥Two incandescent light bulbs draw 60 W and 40 W individually when connected across 240 V DC supply, what are their resistances ?

$\huge\bold{ANSWER}$

✯Difference Between Electric Energy and Electric Power :-

Electric Energy

• The energy supplied by the potential source in a circuit causing the flow of electric current is called electrical energy.

• Mathematically, It is the dot product of Power and Time and Expressed in Joules or Watt-seconds. But it's commercial units for household purpose is Unit. (1unit=1KwH=1000w×3600secs=3.6×10⁶W-sec)

• $e = Pt = VIt = {i}^{2} rt = \dfrac{ {V}^{2}t }{r}$ (Where E is Electric Energy,P is power, V is potential difference, I is electric current , t is time , r is resistance)

Electric Power

• The rate of electric energy is consumed or dissipated by an electrical device in a circuit is called electric power.

• Mathematically, It is Energy Consumed Divided by Time taken. Basically Expressed in Watt unit .

• $p = \dfrac{e}{t} = VI = {I}^{2} r = \dfrac{ {V}^{2} }{r}$ (Where P is power, E is Electric Energy, V is potential difference, I is electric current, t is time , r is resistance)

✯Given:-

2 incandescent light bulbs named $(B_1)$ and $(B_2)$ are individually connected across 240V DC supply.

➜Power drawn by B1=60W$(P_1)$

➜Power drawn by B2=40W$(P_2)$

To Find:-

Resistances of Each Bulb

Solution:-

From Formula We Knew That

$= > p = \dfrac{ {V}^{2} }{r}$

Putting out the values in above equation, we get

$= > 60 = \dfrac{ {240}^{2} }{ r_1} \\ = > r_1 = \frac{240 \times 240}{60} = 960ohm \\ \\ = > 40 = \dfrac{ {240}^{2} }{ r_2} \\ = > r_2 = \frac{240 \times 240}{40} = 1440ohms$

. ˙ .Therefore we get Resistance of 1st Bulb as 960 ohms and Resistance of 2nd Bulb as 1440 ohms

ADDITIONAL INFORMATION

• CHARGE:- The property of matter due which electrostatic force is experienced and Electric field is formed. Represented in C. Unit is Columb

• POTENTIAL DIFFERENCE (V):- The work done to bring an unit positive charge from one point to another point in an electric field is called as the potential difference between the two points. Represented by V. Unit is Volt. $v = \dfrac{ {U} _{2} -U _ 1}{q}$

• ELECTRIC CURRENT:- The rate of amount of charges flowing in electric field is called as Electric current.Unit is Amphere and represented in I.So Mathematically,=> I=Q/T