Question

A transformer with a primary coil of 400 turns and the secondary coil of 100 turns if the primary current is 3A. What is the secondary coil?

Answer 1

Given :

▪ No.of turns in primary coil = 400

▪ No.of turns in secondary coil = 100

▪ Current flow in primary coil = 3A

To Find :

▪ Current flow in secondary coil.

Formula :

➡ The voltage and currents in a transformer are related as

$\underline{\boxed{\bf{\orange{\dfrac{V_S}{V_P}=\dfrac{I_P}{I_S}=\dfrac{N_S}{N_P}=k}}}}$

Calculation :

• Np = 400
• Ns = 100
• Ip = 3A
• Is = (?)

$\implies\bf\:\dfrac{I_P}{I_S}=\dfrac{N_S}{N_P}\\ \\ \implies\sf\:\dfrac{3}{I_S}=\dfrac{100}{400}\\ \\ \implies\sf\:I_S=4\times 3\\ \\ \implies\underline{\boxed{\bf{\gray{I_S=12A}}}}$

____________________________

✴For a step-up transformer, k > 1

✴For a step-down transformer, k < 1

Answer 2

Secondary current is 12Amphere.

Solution

* Number of turns in primary coil = 400

* Number of turns in secondary coil = 100

* Primary current is 3A

* Secondary current = ?

We know the relation b/w voltage, current & number of turns inside a electric circuit.

✪ I_p/I_s = N_s/N_p

Here,

• I_p denotes primary current.
• I_p = 3A

• I_s denotes secondary current.
• I_s = ?

• N_s denotes turns in secondary coil
• N_s = 100

• N_p denotes turns in primary coil
• N_p = 400

Putting in the formula we get :

⇒ 3/I_s = 100/400

⇒ 3/I_s = 1/4

⇒ I_s = 3 × 4

⇒ I_s = 12A

∴ Current in secondary coil = 12 A