Changing galvanometer reading in the secondary coil of a transformer?

[pkc111]

Homework Statement

See photo
How could the size of the deflection be increased?

Relevant Equations

V1/V2 = n1/n2

The present setup is a step-down transformer.

Im thinking if n1 is decreased to match or be less than n2, then it will be a step up transformer.
so A.

(but answers say B)?

 

[Steve4Physics]

Relevant Equations:: V1/V2 = n1/n2

The present setup is a step-down transformer.

Im thinking if n1 is decreased to match or be less than n2, then it will be a step up transformer.
so A.

(but answers say B)?

You are correct (answer A). The official answer (B) is wrong.

 

[pkc111]

Thank you Steve

 

[gneill]

Hmm. A galvanometer is not a voltmeter, it's essentially a current meter. So you want to increase the secondary current in order to increase the deflection. To me it looks like answer (B) is correct.

 

[Steve4Physics]

Hmm. A galvanometer is not a voltmeter, it's essentially a current meter. So you want to increase the secondary current in order to increase the deflection. To me it looks like answer (B) is correct.

Surely the current through the galvanometer is proportional to the voltage applied across it.

Reducing the number of turns (${N_1}$ )of the primary increases the ratio $\frac{N_2}{N_1}$ which increases the induced emf in the secondary coil.

The voltage across the galvanometer is a fraction of this induced emf (the fraction depending on the resistances of the galvanometer and the secondary winding).

 

[gneill]

An ideal galvanometer should be taken to have zero resistance. It responds to the current passing through it. In real-life situations a good galvanometer will have very low resistance.

In an ideal transformer the voltage and current goes as:

$$\frac{n_1}{n_2} = \frac{V_1}{V_2} = \frac{I_2}{I_1}$$

So the larger the "step_down" ratio, the larger the induced current in the secondary.

Edit: Fixed typo in the equation.

 

[Steve4Physics]

An ideal galvanometer should be taken to have zero resistance. It responds to the current passing through it. In real-life situations a good galvanometer will have very low resistance.

In an ideal transformer the voltage and current goes as: $$\frac{n_1}{n_2} = \frac{V_1}{V_2} = \frac{I_2}{I_1}$$
So the larger the "step_down" ratio, the larger the induced current in the secondary.

If we consider a truly ‘ideal’ circuit, i.e. one with all resistances zero, that would lead to zero currents (infinite time constants in primary and secondary inductive circuits).

But I think I agree with you - the question is asking how to increase the secondary current (because a low-resistance galvanometer acts as an ammeter). This gives answer B.

If the diagram showed a voltmeter, the question would be asking how to increase the secondary voltage. This would give answer A.

In my own defence I’ll note that galvanometers typically have resistances in the range 1Ω to 100Ω. It’s not entirely clear if the galvanometer in the question would be acting in ‘ammeter mode’ or ‘voltmeter mode’. So when the question refers to ‘galvanometer deflection’ there is ambiguity whether to consider current or voltage.