Question

1) The rest positions of the needles in a millimeter and voltmeter not in use are as shown
in Fig a. when a student used these in his experiment, the readings of the needle are
in the positions shown in Fig. B. The corrected values of current and voltage in the
experiment are(1M)
a. 34 mA and 4.0 V
b.42 mA and 4.0 V
c. 34 mA and 3.2 V
d. 42 mA and 3.2 V​

Answer 1

Answer:

D

Explanation:

Consider the ammeter.

while not in use it has a negative error of 4.

how i got 4?

see, the readings carefully.. it shows reading 0 and 10. and it has 5 divisions in between. so each division of scale corresponds to (10-0)/5

i.e

(upper division value - lower division value) ÷ (total number of divisions)

So we have, each division corresponds to 2 mA

from fig. the needle is at two divisions below zero. so 2×2 i.e 4mA negative reading is there.

so to get the actual reading we should add this 4 mA

i.e, while the meter is supposed to show zero, it shows -4mA. but to get 0 we added 4mA right?

similarly, as the case B, when the meter shows 38mA, we should add the 4mA , and we get 42mA..

(In case you dont know, how i read the reading as 38mA in case B)

we read 38mA from the fig, since the needle is at 4th division from 30. so 30+ 2×4= 30+8=38

now, voltmeter.

In case A, the meter shows additional or excess reading of 0.4V

(in case you don't know how 0.4V)

the needle is 2 division above 0.

and each division corresponds to (1-0)/5 = 1/5 = 0.2V using the formula mentioned above

so for 2divisions, the reading is 0.4V

So we have excess reading. so to get the actual reading, we have to subtract this excess reading from the reading shown by the meter.

see, when the meter was idle, it showed 0.4V excess and we subtracted 0.4 from 0.4 to get 0

now in case B, the meter is showing 3.6V, and we have to subtract 0.4V from this to get actual reading.

so 3.6 - 0.4= 3.2V is the actual or corrected reading.

(in case you don't know how you read 3.6V in case B)

the needle is above 3 divisions from 3. and we found out each division corresponds to .2V.

so 3 divisions from 3 could be read as

3+ 3 × 0.2 = 3 + 0.6 = 3.6V

so 42mA and 3.6V

i.e, option D is the answer