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It is required to measure the resistance...

It is required to measure the resistance of a circuit operating at `120 V`. There is only one galvanometer of current sensitivity `10^(-6)` A per division. How should the galvanometer be connected in the circuit to operate an ohmmeter? Why minimum resistance can be measured with such a galvanometer if its full-scale has `40` divisions?

Text Solution

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We should connect this galvanometer in series with the gives resistor. In general, resistance of galvanometer is small, so it can be neglected. If `I` is the current in circuit, then `R = 120//I`.
`I_(max) = 40 xx 10^(-6) A, R_(min) = (120)/(40 xx 10^(-6)) = 3 M Omega`
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