2024 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (4) $220\ \text{V},\ 5\ \text{A}$

Given information

This question is about a transformer, which changes an a.c. voltage.

The primary coil has $200$ turns. The secondary coil has $400$ turns, twice as many.

The primary gets $110\ \text{V}$ and carries $10\ \text{A}$. The transformer is ideal, so no energy is lost.

What we must find: the voltage and the current in the secondary.

How we find it: more turns give a bigger voltage, in the same ratio. Then the power stays the same, so the current must get smaller.

$N_P=200$ and $N_S=400$. In the primary, $V_P=110\ \text{V}$ and $I_P=10\ \text{A}$, both r.m.s. values.

What the question asks

$N_P$, $N_S$: numbers of turns in the primary and the secondary.

$V_P$, $V_S$ ($\text{V}$): r.m.s. voltages of the primary and the secondary.

$I_P$, $I_S$ ($\text{A}$): r.m.s. currents in the primary and the secondary.

Relevant theory

If $N_p$ is the number of turns in the primary, $N_s$ is the number of turns in the secondary, $V_P$ is the input voltage and $V_s$ is the output voltage, $$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$

Also if the transformer is an ideal one, the input power supplied from the primary will be output from the secondary without any loss,

Hence, $V_P I_P = V_S I_S$

What is happening: the a.c. in the primary makes a changing magnetic flux in the soft iron core. The same flux passes through every turn of the secondary.

Each turn gets the same share of voltage. So twice the turns give twice the voltage.

No power is lost in an ideal transformer. So when the voltage goes up, the current must go down by the same factor.

For example, think of a long lever. You push one end down a long way with a small force.

The short end rises only a little, but with a big force.

The work you put in equals the work you get out. You trade distance for force.

In words: a transformer trades current for voltage in the same way. Power in equals power out.

Required equations

$\dfrac{V_S}{V_P}=\dfrac{N_S}{N_P}$

$V_PI_P=V_SI_S$

$I_S=\dfrac{V_PI_P}{V_S}$

Applying it to this question

$\dfrac{N_S}{N_P}=\dfrac{400}{200}$

$2$

$\dfrac{N_S}{N_P}$

$V_S=2\times110$

$220$

$V_S$

$1100$

$V_PI_P=110\times10$

$1100=220\times I_S$

$I_S=\dfrac{1100}{220}$

$5$

$I_S$

Reasoning and calculation

The secondary has twice the turns, so the voltage doubles to $220\ \text{V}$. The power stays $1100\ \text{W}$, so the current halves to $5\ \text{A}$. Only option (4) has both.

$55\ \text{V}, 20\ \text{A}$ turns the ratio upside down. The secondary has more turns, so its voltage must be larger than $110\ \text{V}$, not smaller.

$440\ \text{V}, 5\ \text{A}$ multiplies the voltage by $4$, but the turns ratio is only $400/200=2$. The power out would be $440\times5=2200\ \text{W}$, twice the $1100\ \text{W}$ put in.

$220\ \text{V}, 10\ \text{A}$ has the right voltage but keeps the current. The power out would be $220\times10=2200\ \text{W}$. An ideal transformer cannot give out more power than it takes in.

$55\ \text{V}, 10\ \text{A}$ halves the voltage instead of doubling it, and keeps the current. The power out would be $550\ \text{W}$, only half of the $1100\ \text{W}$ put in.

Checks

$110*10/220$

$V_PI_P=110\times10=1100\ \text{W}$

$V_SI_S=220\times5=1100\ \text{W}$

$V_PI_P=V_SI_S$

On both sides $\text{V}\times\text{A}=\text{W}$. The turns ratio has no unit, so $V_S$ keeps the unit $\text{V}$.

The secondary has more turns, so this is a step-up transformer. In a step-up transformer the secondary current is less than the primary current. Here $5\ \text{A}<10\ \text{A}$, as expected.

Conclusion

$V_S=220\ \text{V},\ I_S=5\ \text{A}$

The secondary has $220\ \text{V}$ and $5\ \text{A}$. The answer is (4).