Answer:
a. 60 V b. 6 A c. 360 W
Explanation:
a. Voltage in secondary
For an ideal transformer,
N₁/N₂ = I₂/I₁ = V₁/V₂ where N₁ = turns in primary = 50 turns, N₂ = turns in secondary = 250 turns, V₁ = voltage in primary = 12 V, V₂ = voltage in secondary = ? V
N₁/N₂ = V₁/V₂
V₂ = V₁N₂/N₁ = 250 × 12/50 = 60 V
b. Since V₂ = I₂R,
I₂ = V₂/R R = 10 Ω
I₂ = 60 V/10 Ω
I₂ = 6 A
c. We first calculate the current in the primary from
N₁/N₂ = I₂/I₁ where I₁ = primary current
I₁ = N₂I₂/N₁ = 250 × 6 A/50 = 30 A
The power supplied to the primary is thus
P = I₁V₁ = 30 A × 12 V = 360 W
(a) The voltage supply at the secondary coil is of 60 V.
(b) The current flow through the 10- ohm device is 6 A.
(c) The power supply to the primary coil is 360 W.
Given Data:
Number of turns of primary coil is, n1 = 50.
Number of turns of secondary coil is, n2 = 250.
The voltage supply to primary coil is, V1 = 12 V.
The resistance of device is, R = 10 ohm.
(a)
The first part of the given problem is based on the Transformer equation relating the number of turns of coils at primary and secondary and voltage supply at each coils.
Therefore,
n1 / n2 = V1 / V2
Here,
V2 is the voltage supply to the secondary coil.
Solving as,
50 / 250 = 12 / V2
V2 = 250 / 50 × 12
V2 = 60 V.
Thus, we can conclude that the voltage supply at the secondary coil is of 60 V.
(b)
Now in order to find the current flow through the 10 - ohm device, we can use the Ohm's law as,
V2 = I' × R
Solving as,
60 = I' × 10
I ' = 60 / 10
I' = 6 A
Thus, we can conclude that the current flow through the 10- ohm device is 6 A.
(c)
Now, we first calculate the current in the primary from
n1 / n2 = I' / I
here,
I is the primary current
Solving as,
I = n2 × I' / n1
I = 250 × 6 /50
I = 30 A
The power supplied to the primary is thus
P1 = I × V1
P1 = 30 × 12
P1 = 360 W
Thus, we can conclude that the power supply to the primary coil is 360 W.
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