#### psmao

##### Member

question 3 in this document, shouldnt the answer be A? As $\bg_white \frac{I1}{I2}=\frac{N2}{N1}$

#### jazz519

##### Moderator
Moderator
i would say check hsc answers. I didn’t do any of the questions I just copied and pasted across the answers

#### psmao

##### Member
its a question from their additional sample questions, but i think the answer is wrong or i am totally retarded lol

#### Arrowshaft

##### Active Member
Well as you said, for an ideal transformer
$\bg_white \dfrac{I_s}{I_p}=\dfrac{N_p}{N_s}$
Hence,
$\bg_white I_s=I_p\dfrac{N_p}{N_s}$
Now, for the galvanometer to deflect the most, the current produced has to be the highest, this only happens when $\bg_white I_s$ is big, so either when $\bg_white N_p$ is increased or $\bg_white N_s$ is decreased. Hence the number of secondary coils should be decreased for the largest deflection, so B.

#### Idkwhattoput

##### Member
In this question, I get the correct graph, except it is the negative of what the answers say (answers say initially voltage is, negative but I keep getting positive). I have a feeling it has something to deal with conventional vs actual current. Does anyone know why? The reason I say induced voltage is positive is because induced current runs from the positive terminal to the negative terminal. Maybe that assumption is wrong?

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#### Arrowshaft

##### Active Member
In this question, I get the correct graph, except it is the negative of what the answers say (answers say initially voltage is, negative but I keep getting positive). I have a feeling it has something to deal with conventional vs actual current. Does anyone know why? The reason I say induced voltage is positive is because induced current runs from the positive terminal to the negative terminal. Maybe that assumption is wrong?
Its because the positive current enters at the negative terminal or electrode, hence the electrode measuring ‘negative’ current so to speak, would measure positive current instead.

#### Idkwhattoput

##### Member
Its because the positive current enters at the negative terminal or electrode, hence the electrode measuring ‘negative’ current so to speak, would measure positive current instead.
So the way a voltemeter's terminals works is such that they are arrnaged oppositely to a cell/battery? I ask this because for a cell, 'negative current' ia from negative to positive, and 'postive' current is from positive to negative, but you suggest that for a voltemeter current going from the positive temrinal to the negative is actually negative current.

#### Arrowshaft

##### Active Member
So the way a voltemeter's terminals works is such that they are arrnaged oppositely to a cell/battery? I ask this because for a cell, 'negative current' ia from negative to positive, and 'postive' current is from positive to negative, but you suggest that for a voltemeter current going from the positive temrinal to the negative is actually negative current.
They are not operating as a cell, they simply measure any positive or negative voltage at either end.

#### Idkwhattoput

##### Member
Actually, i get it now i think. The voltemeter's positive and negative terminals are such that they indicate the direction of current going through voltemeter rathwr than around the curcuit