Question

40 lamps, each of resistance 4 ohms are connected in series to 240V supply in order to decorate a tree calculate a) the current in each lamp b) the power dissipated in each lamp show your working and write units​

Answer 1

Given :

• Resistance of Lamp, R = 4 ohm
• No. of lamps = 40
• Potential Difference, V = 240 V

To Find :

• (a) the current in each lamp, l
• (b) the power dissipated in each lamp, P

Solution :

Using Ohm's Law,

• V = IR

Where,

• V is the Potential Difference
• I is the Current
• R is the Resistance

Substitute the value we get,

⇒ 240 = I × 4

⇒ I = 240/4

⇒ I = 60 A

Therefore, the current through the each lamp is 60 Ampere.

Now, Calculating the Power dissipate in each lamp,

• P = VI

Where,

• P is the power
• V is the Potential Difference
• I is the Current

Substitute the value we get,

⇒ P = 240×60

⇒ P = 14400 W

Therefore, the power dissipate through each lamp is 14400 Watts.

Total power dissipate by 40 Lamps = 40 × 14400 = 576000 Watts.

Answer 2

Explanation:

${\large{\bold{\rm{\red{Given \; that}}}}}$

★ Length of rectangle = 17 cm

★ Breadth of rectangle = 13 cm

${\large{\bold{\rm{\green{To \; find}}}}}$

★ Area of rectangle.

${\large{\bold{\rm{\orange{Solution}}}}}$

★ Area of rectangle =

${\large{\bold{\rm{\pink{Using \; concept}}}}}$

★ Area of rectangle formula.

${\large{\bold{\rm{\purple{Using \; formula}}}}}$

★ A = L × B

${\large{\bold{\rm{Where,}}}}$

★ A denotes area

★ L denotes length

★ B denotes breadth

${\large{\bold{\rm{\color{lime}{Full \; solution}}}}}$

★ A = L × B

★ A = 17 × 13

★ A = 221 cm²

${\frak{Henceforth, \: 221 \: cm^{2}}}$

${\Large{\bf{\underbrace{\color{lightgreen}{Additional \; information}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \: \sqrt \dfrac{v}{\pi h}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$