Question

A coil consists of 2000 turns of copper wire having a cross-sectional area of 0.8 mm2. The mean length per turn is 80 cm and the resistivity of copper is 0.02 OHM-m. Find the resistance of the coil and power absorbed by the coil when connected across 110 V d.c. supply.​

Answer 1

Given :-

• Number of Turns = 2000 turns.
• Area of Cross - sectional area = 0.8mm².
• Mean length per turn = 80 cm
• Resistivity of Copper = 0.02 μΩ–m.
• Volt = 110 d.c

To find :-

• Resistance of the coil.
• Power absorbed.

Solution :-

● Length of the coil

$\: \: \: \: \: \: \: \: { \bf{ l = 0.8 \times 2000 }} \\ { \underline{ \boxed{ \bf{ l= 1600m }}}}$

● Cross - sectional Area of Wire is

${ \bf{A = 0.8 \: mm^{2} }}\\\: \: \: \: \: \: \: \: \: \: \:{ \underline{ \boxed{ \bf{A = 0.8 \times 10^{ - 6} {m}^{2} }}}}$

● Resistivity of copper

${ \bf{\rho = 0.02 \: \mu \O-m }} \\ \: \: \: \: \: \: \: \: \: { \bf{= 2 \times 10 ^{ -2} \times 10 ^ {-6} \O-m}} \\ { \underline{ \boxed{ \bf{= 0.02 \times 10 ^{-6} \: \O-m}}}}$

■ Resistance of the coil is :-

${ \large{ \underline{ \boxed{ \bf{ \green{R =\rho \: \cfrac{l }{A }}}}}}}$

${ \large{ \bf{R =\rho \: \cfrac{l }{A }}}}$

${ \bf{R =(0.02 \times 10^{ - 6} ) \times \cfrac{1600 }{(0 .8 \times 10^{ - 6}) }}}$

${\bf{R =0.02 \times \cfrac{1600}{0.8} }}$

${\bf{R = 0.025 \times 1600}}$

$\Large{ \underline{ \boxed{ \red{\bf{R = 40 \: \: Ohm }}}}}$

■ Power absorbed by coil is :-

${ \large{ \underline{ \boxed{ \green{ \bf{Power \: \: absorbed = \frac{ V ^{2} }{R }}}}}}}$

Here

• V = Volt
• R = Resistance

${ \bf{Power \: absorbed = \cfrac{(110)^{2} }{40} }}$

${ \bf{Power \: \: absorbed = \cfrac{110 \times 110}{40} }}$

${ \bf{Power \: \: absorbed = 2.75 \times 110}}$

${ \Large{ \underline{ \boxed{ \red{ \bf{Power \: absorbed =302.5 \: W}}}}}}$

○ Final Answer :-

• Resistance of the coil is 40 Ω
• Power Absorbed is 302.5 W

Answer 2

Answer:

302.5W is correct answer

Explanation:

Given:-

===>no of turn:2000

===>Mean length:80 cm

===>Resistivity of copper: 0 .02 ohm

===>volt:110.