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InterviewSolution
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Two resistors `400Omega` and `800Omega` are connected in series with a `6V` battery. It is desired to measure the current in the circuit. An ammeter of `10Omega` resistance is used for this purpose figure. What will be the reading in the ammeter? Similarly, if a voltmeter of `10,000Omega` resistance is used to measure the potential difference across the `400Omega` resistor, what will be the reading in the voltmeter? |
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Answer» The circuits are shown in figure. In figure, all the resistors are in series, `:.` total resistance of the circuit, `R=400+800+10=1210Omega` Hence, current in the circuit, `I=V/R=(6)/(1210)=0*00496A` Reading in ammeter is `0*00496A` In figure, the resistors of `400Omega` and `10,000Omega` are in parallel, their effective resistance `R_p` will be `R_p=(400xx10,000)/(400+10,000)=(5000)/(13)Omega` `:.` Total resistance of the circuit `=(5000)/(13)+800=(15400)/(13)Omega` Current in the circuit, `I_1=(6)/(15400//13)=(39)/(7700)A` Potential difference across A and B `=I_1R_p=(39)/(7700)xx(5000)/(13)=(150)/(77)=1*95V` `:.` Reading of voltmeter is `1*95V` |
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