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A galvanometer works as 1 V full scale voltmeter when connected in series with resistance of 2kohmand as 500 mA ammeter when connected in parallelwith a resistance of 0.2ohm. Find internal resistanceof galvanometer |
| Answer» L RESISTANCE of galvanometer is 222 ΩLet internal resistance of galvanometer is G.galvanometer is connected with R = 2kΩ in series combination. so, V = (R + G)Ig⇒Ig = V/(R + G) here, V = 1 volt so, Ig = 1/(2000 + G) .......(1)now, an Ammeter is connected in parallel with a resistance of 0.2 Ωso, Ig × G = (500mA - Ig) × 0.2Ω⇒Ig × G = (0.5 A - Ig) × 0.2 ⇒(0.2 + G) × Ig = 0.1 ⇒(0.2 + G) × 1/(2000 + G) = 0.1 [ from eq (1) ]⇒(0.2 + G) = 200 + 0.1G⇒0.9G = 199.8 ⇒G ≈ 222Ωhence, internal resistance of galvanometer is 222 Ωalso read similar questions: A galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1.5 vol... brainly.in/question/9756087A galvanometer of resistance G is CONVERTED into a voltmeter to measure upto V volts by connecting a resistance R1 in SE... brainly.in/question/8007945 | |