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Explain : "Increasing the current sensitivity may not necessarily increase the voltage sensitivity". |
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Answer» Solution :1. CURRENT sensitivity is given by, `phi/I=(NAB)/k""...(1)` 2. If N `to` 2N that is we double the number of turns, then `phi/I=(2phi)/I` 3. Thus, the current sensitivity doubles. HOWEVER the resistance of the galvanometer is LIKELY to double since it is proportional the length of the wire. 4. Voltage sensitivity, `phi/V=((NAB)/k)(1/V)=((NAB)/k)1/R` If `Nto2NandRto2R` then, voltage sensitivity does not change. `phi/Vto phi/V` 5. So, it is not NECESSARY to change in voltage sensitivity by changing current sensitivity. |
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