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Question: A micro-ammeter has a resistance of 100 ohms and a full-scale range of 50 micro-amperes. It can be u...

A micro-ammeter has a resistance of 100 ohms and a full-scale range of 50 micro-amperes. It can be used as a voltmeter or as a higher range ammeter provided resistance is added to it. Pick the current range and the resistance combinations.
A. 50 Volts range with 10KΩ10K\Omega resistance in series
B. 10 Volts range with 200KΩ200K\Omega resistance in series
C. 5mA range with 1Ω1\Omega resistance in parallel.
D. 10mA range with 1Ω1\Omega resistance in parallel.

Explanation

Solution

In this question, we need to determine the extended range of the meter and the value of the resistance added with the initial value of the resistance. For this, we will use the concept of Kirchhoff’s law. Here, we need to check all the given options and satisfy the condition.

Complete step by step answer: An ammeter can act as a voltmeter if it is connected in series with the load along with the high resistance.
For the micro-ammeter to measure the extended current in the circuit, we will use the formula S=RgIgI−IgS = \dfrac{{{R_g}{I_g}}}{{I - {I_g}}} where, Rg{R_g} is the internal resistance of the ammeter, Ig{I_g} if the current flowing through the meter, S is the additional parallel resistance connected to the ammeter, and I is the extended value of the current.

Following the ohm’s law, which states that the product of the equivalent resistance and current results in the voltage of the given circuit. Mathematically, V=iReqV = i{R_{eq}}
Now, checking the options
(A)For the ammeter to behave as a voltmeter to measure 50 volts of potential difference.
V=iReq 50=(50μA)(100+R) (100+R)=5050×10−6 100+R=106 R≈100KΩ  V = i{R_{eq}} \\\ 50 = \left( {50\mu A} \right)\left( {100 + R} \right) \\\ \left( {100 + R} \right) = \dfrac{{50}}{{50 \times {{10}^{ - 6}}}} \\\ 100 + R = {10^6} \\\ R \approx 100K\Omega \\\
But in the option, it is given as 10KΩ10K\Omega so, option A is not correct.
(B)For the ammeter to behave as a voltmeter to measure 10 volts of potential difference.
V=iReq 10=(50μA)(100+R) (100+R)=1050×10−6 100+R=0.2×106 R≈200KΩ  V = i{R_{eq}} \\\ 10 = \left( {50\mu A} \right)\left( {100 + R} \right) \\\ \left( {100 + R} \right) = \dfrac{{10}}{{50 \times {{10}^{ - 6}}}} \\\ 100 + R = 0.2 \times {10^6} \\\ R \approx 200K\Omega \\\
Hence, option B is correct.
(C) For the micro-ammeter to measure the extended current in the circuit, we will use the formula S=RgIgI−IgS = \dfrac{{{R_g}{I_g}}}{{I - {I_g}}} as:
S=RgIgI−Ig =100×(50μA)5mA−(50μA) =100×50×10−65×10−3−50×10−6 =5×10−35×10−3−0.05×10−3 =54.95 =1.01Ω≈1Ω  S = \dfrac{{{R_g}{I_g}}}{{I - {I_g}}} \\\ = \dfrac{{100 \times \left( {50\mu A} \right)}}{{5mA - \left( {50\mu A} \right)}} \\\ = \dfrac{{100 \times 50 \times {{10}^{ - 6}}}}{{5 \times {{10}^{ - 3}} - 50 \times {{10}^{ - 6}}}} \\\ = \dfrac{{5 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 3}} - 0.05 \times {{10}^{ - 3}}}} \\\ = \dfrac{5}{{4.95}} \\\ = 1.01\Omega \approx 1\Omega \\\
Hence, option C is correct
(D) ) For the micro-ammeter to measure the extended current in the circuit, we will use the formula S=RgIgI−IgS = \dfrac{{{R_g}{I_g}}}{{I - {I_g}}} as:
S=RgIgI−Ig =100×(50μA)10mA−(50μA) =100×50×10−610×10−3−50×10−6 =5×10−310×10−3−0.05×10−3 =59.95 =0.5025Ω  S = \dfrac{{{R_g}{I_g}}}{{I - {I_g}}} \\\ = \dfrac{{100 \times \left( {50\mu A} \right)}}{{10mA - \left( {50\mu A} \right)}} \\\ = \dfrac{{100 \times 50 \times {{10}^{ - 6}}}}{{10 \times {{10}^{ - 3}} - 50 \times {{10}^{ - 6}}}} \\\ = \dfrac{{5 \times {{10}^{ - 3}}}}{{10 \times {{10}^{ - 3}} - 0.05 \times {{10}^{ - 3}}}} \\\ = \dfrac{5}{{9.95}} \\\ = 0.5025\Omega \\\
Hence option D is incorrect
Therefore, options B and C are correct.

Note: Students should be aware of the placement of the internal resistance of the voltmeter and the ammeter, as well as the placement of additional resistances that should be connected with the internal resistances of the meters to extend the range.