Question

The turn ratio of a transformer is given as 2:3. If the current through the primary coil is 6 A, thus calculate the current through load resistance?

• A. 4 A
• B. 4.5 A
• C. 2 A
• D. 1.5 A

Answer 1

Answer: (A)

4 A

Answer 2

For an ideal transformer, the ratio of the number of turns in the primary coil ($N_p$) to the number of turns in the secondary coil ($N_s$) is related to the ratio of the currents in the primary coil ($I_p$) and the secondary coil ($I_s$) by the formula:

$\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$

In this question, the turn ratio of the transformer is given as 2:3. This means the ratio of the number of turns in the primary coil to the number of turns in the secondary coil is $N_p : N_s = 2 : 3$. So, $\dfrac{N_p}{N_s} = \dfrac{2}{3}$.

The current through the primary coil is given as $I_p = 6$ A. We need to calculate the current through the load resistance, which is the current through the secondary coil, $I_s$.

Using the formula $\dfrac{I_s}{I_p} = \dfrac{N_p}{N_s}$, we can substitute the given values: $\dfrac{I_s}{6 \text{ A}} = \dfrac{2}{3}$

Now, we solve for $I_s$: $I_s = 6 \text{ A} \times \dfrac{2}{3}$ $I_s = \dfrac{6 \times 2}{3} \text{ A}$ $I_s = \dfrac{12}{3} \text{ A}$ $I_s = 4 \text{ A}$

The current through the load resistance is 4 A.