# Voltage and loops of transformers – problems and solution

1.< /p>

Based on the above table, find R and P.

Solution

Transformer’s equation :

V_{s }N_{p} = N_{s }V_{p }dan V_{p} I_{p} = V_{s} I_{s}

*V*_{s}* = secondary voltage, V*_{p}* = primary voltage**, N*_{s}* = secondary loops, N*_{p}* = primary loops*_{, }*I*_{p}* = primary current, I*_{s}* = secondary current*

__Primary voltage (V___{p}__)__

V_{p} I_{p} = V_{s} I_{s}

V_{p} (4) = (10)(80)

V_{p} = 800 / 4

V_{p} = R = 200 Volt

__Secondary loops (N___{s}__)__

V_{s }N_{p} = N_{s }V_{p}

(10)(600) = N_{s }(200)

6000 = 200 N_{s}

N_{s }= 6000 / 200

N_{s} = P = 30 loops

2.

Based on the above table, find K and L.

Solution :

__Secondary current (I___{s}__) of transformer 1 :__

V_{p} I_{p} = V_{s} I_{s}

(200)(3) = (300)(I_{s})

I_{s} = 600 / 300

I_{s} = K = 2 A

__Secondary voltage (V___{s}__) of transformer 2 :__

V_{p} I_{p} = V_{s} I_{s}

(40)(0,75) = V_{s} (1)

V_{s} = L = 30 Volt

3.

Based on the table above, find P and Q.

Solution :

__Primary voltage (V___{p}__)__

V_{s }N_{p} = N_{s }V_{p}

(220)(300) = (600) V_{p}

(220)(3) = (6) V_{p}

660 = 6 V_{p}

V_{p} = 660 / 6

V_{p} = P = 110 Volt

__Primary current (I___{p}__)__

V_{p} I_{p} = V_{s} I_{s}

(110)(I_{p}) = (220)(2)

(110)(I_{p}) = 440

I_{p} = 440 / 110

I_{p} = Q = 4 A