For the first one 320
second
1200W
Data
R = 12 Ω ∆V = 120V I =? P =?
Solution:
According to Ohm’s law,
∆V = I R
I = ∆V / R
= 120 / 12
= 10 A
Power P = I ∆V
= 10 x 120
= 1200 W
Third
∆V = 120 V P = 60 W I =? R =?
Use the formula, P = I ∆V
I = P / ∆V = 60 / 120 = 0.5 A
∆V = I R
R = ∆V / I = 120 / 0.5 = 240 Ω
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Answer:
The answer is below
Explanation:
Given that:
The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²
the relative permittivity of dielectric (εr) is 7.0
Permittivity of free space (εo) = 8.854 × 10^(-12)
capacitance of 100uF
potential difference (V) of 12V
d = separation between plate
The capacitance (C) of a capacitor is given by:

The electric field between plates is given as:
E = V /d

Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for
DAC:

⇒ 
Now for the
BAC:

⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle,
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 

After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x = 
Answer:
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Explanation: