InterviewSolution
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InterviewSolution
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A galvanometer reads `5*0V` at full scale deflection and is graded according to its resistance per volt at full scale deflection as `5000OmegaV^-1`. (i) How will you convert it into a voltmeter that reads `20V` at full scale deflection? (ii) Will it still be graded `5000OmegaV^-1`? (iii) Will you prefer this voltmeter to one that is graded `2000OmegaV^-1`? |
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Answer» Here, resistance per volt=`5000OmegaV^-1`. It means for `1` volt pot. difference, the resistance of galvanometer is `5000Omega`. So current for full scale deflection, `I_g=1/5000A=0*2xx10^-3A` (i) In order to convert it into a voltmeter and range 0 to `20V`, let a resistance R be connected in series with it. Then on applying an extra potential difference `=20V-5V=15V`, the current potential it for full scale deflection is again `0*2xx10^-3A` i.e., `I_g=0*2xx10^-3A`. `:. RI_g=15` or `R=15/I_g=(15)/(0*2xx10^-3)=75000Omega` Thus, to convert a given voltmeter (of range 0 to 5V) into a voltmeter (of range 0 to 20V) a resistance of `75000Omega` should be connected in series. With the given meter. (ii) Original resistance of voltmeter `=5000OmegaV^-1xx5V=25000Omega` Total resistance after conversion `=25000+75000=100,000Omega` Resistance per volt of new meter `=(100,000)/(20)=5000OmegaV^-1` i.e. it has the same grading as before. (iii) If greater is the resistance per volt of a meter, the lesser will be the current drawn from the circuit by it. Due to it, it works better. That is why this meter (graded `5000OmegaV^-1`) is more accurate that the one graded as `2000OmegaV^-1`. |
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