Answer:
a ) 3 v
b ) 400 mA
Explanation:
In case of transformer there is relation between no of turns in primary and secondary coils and voltage input and output as follows
V₁ / V₂ = n₁ / n₂
Where V₁ and V₂ are input and output voltage and n₁ and n₂ are no of turns of primary and secondary coils.
Here V₁ = 1.5 v , V₂ = ? , n₁ = 10 , n₂ = 20
Put the values in the equation above
1.5 / V₂ = 10 / 20
V₂ = 3 v.
Similarly formula for current is as follows
I₁ / I₂ = n₂ / n₁
Whereas I₁ and I₂ are current in primary and secondary coil
Put the values in the equation above
I₁ / 200 mA = 20 / 10
I₁ = 400 mA.
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I think it's 1.03412969 or 1.03
Answer:
a) 4.94e9 J b) 1.07e10 J
Explanation:
The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

a) For the original capacitor, we can find directly U as follows:
U = 4.94*10⁹ J
b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.
As the capacitance is proportional to the dielectric constant, we can write the following proportion:
ε₂ / ε₁ = 

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

⇒ U = 1.07*10¹⁰ J