The science and technology team of a school made a self made ohmmeter that can measure the resistanc

1) What is the indication number that should be marked on the disc at the position of the full deviation of the ohmmeter pointer?

Answer: When the 0A and B terminals are connected together, there is no resistance between the A and B terminals, and the indicator should of course be marked with 02) When A and B are disconnected, what is the indication on the disc at the position pointed by the ohmmeter pointer?

Answer: Infinity.

When A and B are intermittently disconnected, the current cannot pass through, and the resistance is infinite, of course, it should be marked as such.

3) What is the number that should be marked in the middle of the dial of this ohmic? Briefly explain why?

Answer: 15 ;

Reason: The ohmic dial is in the middle of the position, the current in the ammeter is half of the pointer when the needle is full of deviation, by Ohm's law i = u r, when you are unchanged, the current is halved, it should be due to the resistance doubled, that is, the total resistance in the circuit is 2 times the original, and the original total resistance is 15, so the resistance connected between a and b is also 15

4) Is the scale on the ohmic dial uniform? Please explain the reason in terms of the circuit principle.

A: Ungenerosity;

Reason: If it is proportional, please look at the illustration, proportional image, the ordinate varies uniformly with the abscissa, so the scale should be uniform, but from Ohm's law I = u r total + rr, i is not proportional to rr, please look at the illustration, proportional image, every time the abscissa increases by one grid, the number of vertical axis increases is changing, not uniformly increasing, so the scale of the ohm table is uneven.

infinity; 15, because when the pointer is in the middle, the current is half of the full deviation, then the total resistance in the circuit at this time is 2 times the full time, that is, 30, so the position in the middle of the dial should be marked 15;

uneven. Let the total resistance of the internal resistance of the ammeter, the effective value of the sliding rheostat and the fixed resistance R0 be the total resistance of r (15), and let the connected resistance to be measured be Rx. It can be seen from the uniform scale of the dial of the ammeter:

The angle at which the ammeter pointer is deflected is proportional to the current passing through. If the current is also proportional to the resistance to be measured, the scale of the ohmmeter should also be uniform. But according to the circuit there are:

i=U (rr+rx), it can be seen that i is not proportional to rx, so the ohmic scale is uneven.

Hope that helps=w=

The shorting of the pen and the external resistance is 0

2. Infinity Pen Open External Resistance Infinity.

3. According to Ohm's law of the whole circuit, e=i(r+r), when the full deviation is e=i*15, when the middle is e=i(15+r), the solution is x=15 ohms, and the middle should be marked with 15 ohms.

4. The dial scale is uneven The scale is the position of the pointer The magnitude of the current i=e (15+r) It can be seen that the position of the pointer is about inversely proportional to the external resistance It is not a linear relationship and it is uneven.

Solution: (1) When AB is directly connected together, the resistance between AB is zero, full deviation, and this position should be marked.

2) When AB is disconnected, the resistance between AB is infinite, and the indication is the smallest (zero position), and this position should be marked with the resistance value (infinity).

3) When AB is connected together, the full bias current is i, at this time.

When ab is indirectly r', the current is , at this time.

The resistance value should be marked at the dial**.

4) When AB is indirectly resistant, the current, at this time.

The resistance value should be marked at the dial**.

Calculated from the above, the ohmic dial scale corresponding to the current on the sensitive galvanometer dial. As you can see in the figure below, the ohmic dial scale is uneven.

Solution: (1) When AB is directly connected together, the resistance between AB is zero, so the indication number that should be marked on the disk at the full deviation position is zero

2) When AB is disconnected, the resistance between AB is infinite, and the indication is the smallest (zero position), and this position should be marked with the resistance value (infinity).

3) The position in the middle of the dial should be marked with the number 15

Because when the hand is in the middle of the dial, the current in the ammeter is half when the hand is full and off is half of it, which is determined by Ohm's law i=u r

It can be seen that when U is constant, the current is halved, so the total resistance in the circuit at this time should be 2 times the resistance when the full deviation is full, and the original total resistance is 15, so the resistance connected between AB is also 15

4) The scale value on the dial is uneven

Reason: The total resistance value of the internal resistance of the ammeter, the effective value of the sliding rheostat and the fixed resistance R0 is R total (15), and the resistance to be measured is Rx; It can be seen from the uniform scale of the dial of the ammeter: the deflection angle of the ammeter needle is proportional to the passing current; If the current i is also proportional to the resistance rx to be measured, then the scale of the ohmmeter should also be uniform but according to the circuit there are:

i=U (rr+rx), it can be seen that i is not proportional to rx, so the ohmic scale is uneven.

2. The power supply voltage remains unchanged, the circuit resistance increases by 80 ohms, and the current drops by half, according to U IR, the circuit resistance can be doubled, so R1 80 ohms.

3. The power supply voltage is U IR volt.

Total resistance r total u i 48 ohms.

r r total - r 120-80 40 ohms.

The "full scale" line of the ammeter is the "0" of the ohmmeter;

2. The "0 scale" line of the ammeter is the "ohmmeter".

Solution: (2) Let the power supply voltage be u, which is obtained by Ohm's law: i1=ur1

i.e.: ,i2=

ur1+r i.e.: ur1+80, which is solved by u=48v, r1=80 (3) when the current i3=, i3=u

r1+r′48v

80 +r, solution: r = 40 ;

4) From the inscription, the zero tick mark of the ohmmeter is at the far right, and the zero tick mark of the ammeter is at the far left;

Current flowing through the ohmmeter i=

ur1+r, r=ui

r1, it can be seen that r is not proportional to i, and the ohmic scale is uneven, so the answer is: (2) 80; （3）40；(4) The zero tick mark of the ohmmeter is at the right end; Ohmmeter tick marks are uneven

.(2) 80 (3) 40 (4) The scale is uneven, the indication is small on the right and large on the left, and the indication is sparse on the right and dense on the left.

(2) Let the power supply voltage be u, which is obtained by Ohm's law: i1=u

r, i.e., r---i2=ur+r

i.e.: r+60

--Obtained by solution: u=36v, r1=60

3) When the current i3=, i3=u

r+r′36v

60 +r, solution: r = 120 ;

4) From the inscription, the zero tick mark of the ohmmeter is at the far right, and the zero tick mark of the ammeter is at the far left;

The current flowing through the ohmmeter i=your+r

Then r = ui-r1, so it can be seen that r is not proportional to i, and the ohmic scale is not uniform so the answer is:

(4) The zero tick mark of the ohmmeter is at the right end; Ohmmeter tick marks are uneven

1) A and B are short-circuited, the pointer is fully biased, the current is maximum, and the external resistance is 0, then the number should be marked as "0". (2) A and B are disconnected at two points, and the current is 0, and the number pointed by the pointer should be marked as "Cha Chenzhong". 3) According to the complete Ohm's law e=i(r+r), when the full deviation is e=i*15, there is i'defeat in the middle = 1 2i, e=i'(15+rx), the solution is rx=15 ohms, and the middle should be marked with 15 ohms.

4) The dial scale is uneven, because the scale is the position of the pointer The size of the current i=e (15+r) It can be seen that the position of the pointer is inversely proportional to the external resistance, not a linear relationship, and uneven. Hope it helps!

1) The pen is shorted, indicating that the external resistance is zero, so the position is marked "0".

2) The table pen is disconnected, which is equivalent to the infinity of the external resistance, so the position is marked "".

3) We can call "the total resistance of the internal resistance of the sensitive ammeter, the effective value of the sliding rheostat and the fixed resistance R0 is 15" as the ohmic internal resistance, which is represented by R. When the pointer points to the dial, it means that the current is half a scattered half of the full partial current, which means that the total resistance in the circuit becomes 2 times (according to Ohm's law of closed circuits), so at this time, the external resistance (it can also be said that the resistance to be measured) should be equal to the internal resistance r of the ohmmeter, that is, this position is marked 15. Commonly known as median resistance.

4) Uneven Each i value corresponds to each resistance value, and by i = e (r + r outside), it is obvious that i and r are not uniformly changed, so the scale is not uniform.

1) When AB is directly connected together, the resistance between AB is zero, so the indication number that should be marked on the disk at the full deviation position is zero

2) When AB is disconnected, the resistance between AB is infinite, and the indication is the smallest (zero position), and this position should be marked with the resistance value (infinity).

3) The position in the middle of the dial should be marked with the number 15

Because when the hand is in the middle of the dial, the current in the ammeter is half of the time when the hand is full, which is determined by Ohm's law i = <>

It can be seen that when U is constant, the current is halved, so the total resistance in the circuit at this time should be 2 times the resistance when the full deviation is full, and the original total resistance is 15, so the resistance connected between AB is also 15

4) The scale value on the dial is uneven

Reason: Set the effective value of the internal resistance of the ammeter, the sliding rheostat and the fixed resistance r0

The total resistance is r-total.

15) Set the resistance to be measured as rx

It can be seen from the uniform scale of the dial of the ammeter: the deflection angle of the ammeter needle is proportional to the passing current; If the current i is also proportional to the resistance rx to be measured, then the scale of the ohmmeter should also be uniform but according to the circuit there are: i = <>

Visible i and rx

It is not proportional, so the ohmic scale is uneven

1) Connect the wire between A and B, adjust the sliding rheostat R1 to make the current representation number the full scale), and mark the "scale mark" of the ammeter as "0"; rab=0,r1+rab=u (supply voltage) i.e. full scale); When Rab=80, R1+80=U (power supply voltage) The power supply voltage is constant, and the total resistance R1+Rab is inversely proportional to I, yielding: R1=80. （r1+rab）：

160=::4,R1+Rab=120 ,Rab=40 (2) Keep the sliding rheostat R1 unchanged, when the resistance value between AB is 80 resistor, the current is expressed as . At this time, the effective resistance value of the sliding rheostat R1 is 80, and the value of "80" is marked at the scale of the ammeter;

3) According to this principle, the ammeter scale can be changed to the corresponding resistance value through calculation, then the resistance value that should be marked at the ammeter scale is "40;

4) The embedding ohm meter can be used to directly measure the resistance value of the conductor connected to the AB.

Please write down the two differences between the ohmmeter scale and the ammeter scale made by this method: 1. The "full scale" line of the ammeter is the "0" of the ohmmeter; 2. The "0 scale" line of the ammeter is the "ohmmeter".

(2) Let the power supply voltage be u, which is obtained by Ohm's law: i1=u

r, i.e., r---i2=ur+r

i.e.: r+80

--Obtained by solution: u=48v, r1=80

3) According to the title, the range of the ammeter is 0, the graduation value is, and the indication is, when the current i3=, i3=u

r+r′48v

80 +r, solution: r = 400 ;

4) From the inscription, the zero tick mark of the ohmmeter is at the far right, and the zero tick mark of the ammeter is at the far left;

The current flowing through the ohmmeter i=your+r

Then r = ui-r1, so it can be seen that r is not proportional to i, and the ohmic scale is uneven so the answer is: (2) 80; （3）400；(4) The scale is large on the left and small on the right (0 at the right end); The scale is uneven (dense on the left and sparse on the right).