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A potential difference of 220 V is maintained across a 12000 ohm rheostat AB. The voltmeter V has a resistance of 6000 ohm and point C is at one fourth of the distance from A to B. What is the reading in the voltmeter? |
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Answer» Solution :As in CASE of linear rheostat `RpropL,(R_(AC))/(R_(AB))=(AC)/(AB)` Here, as `R_(AB)=12000Omega` and `AC=1/4AB` So, `R_(AC)=12000xx1/4=3000Omega` Now as the resistance `R_(AC) ( = 3000 OMEGA)` is in parallel with VOLTMETER of resistance of `6000Omega`. So the effective resistance between points A and C will be `R_(AC).=(3000xx6000)/((3000+6000))=2000Omega` Now as `R_(BC)=(R_(AB)-R_(AC))` `=12000-3000)=9000Omega` is in SERIES with `R_(AC). (= 2000Omega)` and in series potential divides in proportion to resistance, potential difference ACROSS AC, i.e., voltmeter reading will be, `V_(AC)=(R_(AC).V_(AB))/((R_(BC)+R_(AC).))=(2000)/((9000+2000))xx220=40V` 
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