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Question

A galvanometer of resistance 10ohm and maximum current of2μA is converted into voltmeter of range 100mV and when converted into ammeter then range is 1mA. When these voltmeter and ammeter are connected by a (ideal) battery in series with a resistance ofR=1000Ω, then


A

Measured value of R is between978Ωand996Ω

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B

If the ideal battery is replaced by non-ideal battery with internal resistance of 5Ω then R will be >1000Ω

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C

Shunt resistance is 20mΩ

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D

Resistance of voltmeter 105Ω

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Solution

The correct option is C

Shunt resistance is 20mΩ


Step 1. Given data:

Maximum current flowing through Galvanometer, Ig=2μA,

Ig=2×10-6A

Range of the voltmeter, V=100mV,

V=100×10-3V

Range of Ammeter, I=1mA,

I=1×10-3A

Resistance of Galvanometer, Rg=10Ω ,

Additional resistance connected, R=1000Ω,

Rv= Resistance of Voltmeter

RA= Resistance of Ammeter

Explanation for the correct options

Step 2. Check Option C, if it is correct:

As an ammeter,

Resistance of ammeter RA=G×RsG+Rs
i=igRs+GRs

i=igrgRs
from the given data,
1mA=2×10610Rs

Rs=0.02Ω

Rs=20mΩ
Hence option (B) is correct.

Step 3. Check Option A, if it is correct:

For option A,


Req=980.48Ω

I=V0980.48 and V=V0×50000980.48×51

R(measured)=VI=5000051=980.4Ω

Where the measured value of R is between978Ωand996Ω

Hence option (A) is correct.

Explanation for the incorrect options:

Step 4. Check Option D, if it is correct:
As a voltmeter V=ig(rg+R0)(where, R0= shunt resistance)


reading of voltmeter =100×103
so, 100×103=2×106(10+R0)

R0=50kΩ
Hence option (D) is wrong.

Step 5. Check Option B, if it is correct:

Thus, The maximum value of resistance can not be greater than 1000Ω. Internal resistance of cell will not affect the total resistance measurement.

Hence, option B is incorrect.

Hence, the correct option are A,C.


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