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Two resistances of 100 ohm and 200 ohm are connected in series with a battery of emf 4 volt and negligible internal resistance. A voltmeter of resistance 200 ohm is used to measure voltage across the two resistances separately. Calculate the voltage indicated. |
Answer» Solution : When the voltmeter is connected across `R_(1) (= 100Omega), 100 Omega` and voltmeter RESISTANCE `200Omega` will be in parallel resulting in `R_(1).=(R_(1)xxr)/((R_(1)+r))=(100xx200)/((100+200))=200/3Omega` Now, in the circuit this `R_(1).` is in series with `R_(2) ( = 200 Omega)` and in series potential divides in PROPORTION to resistance so potential difference across `R_(1).`, i.e., either `R_(1)` or voltmeter (as the two are in parallel) will be `V_(1).=(R_(1).)/((R_(1).+R_(2)))V=((200//3))/((200//3)+200)xx4=1` VOLT Similarly, if voltmeter is connected across `R_(2)(= 200Omega)` `R_(2).=(R_(2)xxr)/((R_(2)xxr))=(200xx200)/((200+200))=100Omega` So that, `V_(2).=(R_(2).)/((R_(2).+R_(1)))V=100/((100+100))xx4=2` volt |
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