When a current of 6 A is flowing through the primary coil of an ideal transformer having power of 60 W, output voltage is 12 V. Select the correct answer which gives the type of the transformer and the current ratio (Primary current : Secondary current).

2019 A/L Physics Past Paper · Question 19 (English Medium)

An ideal transformer: its type and the current ratio

When a current of $6\ \text{A}$ is flowing through the primary coil of an ideal transformer having power of $60\ \text{W}$ , output voltage is $12\ \text{V}$ . Select the correct answer which gives the type of the transformer and the current ratio (Primary current : Secondary current).

(1) Step down and 6 : 5
(2) Step down and 5 : 6
(3) Step up and 1 : 2
(4) Step up and 5 : 6
(5) Step up and 6 : 5

Answer and explanation

Answer: (5) Step up and 6 : 5

Given information

Given:

An ideal transformer loses no power.

The power is $P = 60$ W.

The primary current is $I_p = 6$ A.

The output voltage is $V_s = 12$ V.

What the question asks

We want:

The type of the transformer, step up or step down.

The current ratio $I_p : I_s$ .

Relevant theory

In an ideal transformer no power is lost.

The power in is the power out.

$V_pI_p = V_sI_s = P$

If $V_s$ is more than $V_p$ it is a step up transformer.

If $V_s$ is less than $V_p$ it is a step down transformer.

Thus in a step up transformer the current flowing in the secondary is less than that in the primary.

Required equations

$V_pI_p = V_sI_s = P$

$\frac{I_p}{I_s} = \frac{V_s}{V_p}$

Applying it to this question

The plan:

First find $V_p$ from the power.

Compare $V_p$ with $V_s$ for the type.

Then find $I_s$ from the power.

Last, the ratio $I_p : I_s$ .

Reasoning and calculation

$10$

$V_p = \frac{P}{I_p} = \frac{60}{6}$

$5$

$I_s = \frac{P}{V_s} = \frac{60}{12}$

$1.2$

$I_p : I_s = 6 : 5$

$\text{step-up}$

Checks

Unit check:

$\dfrac{\text{W}}{\text{A}} = \text{V}$

$\dfrac{\text{W}}{\text{V}} = \text{A}$

Every value has the right unit.

Does it make sense?

The voltage goes up from 10 V to 12 V.

The current goes down from 6 A to 5 A.

The power in: $10 \times 6 = 60$ W.

The power out: $12 \times 5 = 60$ W.

They are equal.

$Ip / (P / Vs)$

Conclusion

$I_p : I_s = 6 : 5$

$V_p = 10$ V and $V_s = 12$ V.

The voltage goes up. So it is a step up transformer.

$I_s = 5$ A, so $I_p : I_s = 6 : 5$ .

The answer is option (5).

Thinking that the current also goes up in a step up transformer.

The power does not change.

Where the voltage goes up, the current goes down in the same ratio.

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