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andrezito [222]
3 years ago
5

The resistances of the primary and secondary coils of a transformer are 56 and 14 , respectively. Both coils are made from leng

ths of the same copper wire. The circular turns of each coil have the same diameter. Find the turns ratio Ns /Np.
Physics
1 answer:
pshichka [43]3 years ago
6 0

Answer:

\dfrac{14}{56}

Explanation:

Given,

Resistance of the primary coil, R_p = 56 Ω

Resistance of the secondary coil, R_s = 14 Ω

We know,

Number of winding in the transformer is directly proportional to the length of the wire.

If the length of the wire increase the resistance also increase.

So, we can say that Number of turns is directly proportional to the resistance of the wire.

So,

\dfrac{N_s}{N_p}=\dfrac{R_s}{R_p}

\dfrac{N_s}{N_p}=\dfrac{14}{56}

The turn ratio is equal to \dfrac{14}{56}

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A +35 µC point charge is placed 46 cm from an identical +35 µC charge. How much work would be required to move a +0.50 µC test c
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Answer:

512.5 mJ

Explanation:

Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.

The electric potential at this point due to the two charges q is thus

V = kq/r₂ + kq/r₂

= 2kq/r₂

= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m

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= 2739.13 × 10³ V

= 2.739 MV

When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.

So, the new electric potential at this point is

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= kq(1/r₃ + 1/r₄)

= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)

= 315 × 10³(2.857 + 9.091) V

= 315 × 10³ (11.948) V

= 3763.62 × 10³ V

= 3.764 MV

Now, the work done in moving the charge q' to the point 12 cm from either charge is

W = q'(V' - V)

= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)

= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V

= 0.5125 J

= 512.5 mJ

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