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Q.

A simple ohmmeter consists of battery connected in series with galvanometer and resistor, as shown in the figure. The resistance RS is chosen such that when the terminals a and b are shorted (put in electrical contact with negligible resistance between them), the current through the galvanometer gives a full scale deflection. Thus, a full scale deflection indicates no resistance between the terminals a and b. A zero deflection indicates an infinite resistance between the terminals.  When the terminals are connected across an unknown resistance R, the current through the galvanometer depends on R., so the scale can be calibrated to give a direct reading of R, as shown in the figure. Because an ohmmeter sends current through the resistance to be measured, some caution must be exercised when using this instrument. For example, you would not want to try measure the resistance of sensitive ammeter with an ohmmeter. Because the current provided by the battery in the ohmmeter would probably damage the ammeter. Let us use a galvanometer with resistance of 20Ω and maximum current of  10mA, ε=1V  . For a current more than 10mA, galvanometer would be damaged

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The minimum resistance in ohm required for rest of the circuit (other than galvanometer) not to get galvanometer damaged is (10x) then  x=

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answer is 8.

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Detailed Solution

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10×103=120+RR=80Ω

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