Question

A current of 2mA was passed through an unknown resister which dissipated a power of 4.4W . dissipated power when an ideal power supply of 11V is connected across it is ​

Answer 1

Answer:

11×10 −5W

P=I

2

R

4.4=4×10

−6

R

R=1.1×10

6

Ω

P

1

=

R

11

2

=

1.1

11

2

×10

−6

=11×10

−5

W

Answer 2

Answer:

Dissipated power when an ideal power supply of 11c is connected across it is 11×10⁻⁵

Explanation:

power dissipated by any resistor R ,

when I current flows through it is ,

P = i²R -------(1)

• Given , I=2mA = 2× 10⁻²A
• and p = 4.4W

Using equation (1) we get

• 4.4 = (2×10⁻³)² ×R

or, R = 4/4×10⁻⁶ Ω _______(2)

When this resistance R is connected with 11V supply then power dissipated is

• P = v²/R

P = 11²/4.4× 4× 10⁻⁶ [ ∵ using equation (ii) ]

=> P = 11 × 11 × 4 × 10⁻ ⁶ / 44× 10⁻¹ W

• or, P = 11 × 10⁻⁵ W Answer