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Question Number 41434 by rahul 19 last updated on 07/Aug/18

Commented by rahul 19 last updated on 07/Aug/18

If the ammeter A_(2 ) reads 1.6A and ammeter A_3 reads 0.4 A . Find reading of ammeter A_1 ?

$$\mathrm{If}\:\mathrm{the}\:\mathrm{ammeter}\:\mathrm{A}_{\mathrm{2}\:} \:\mathrm{reads}\:\mathrm{1}.\mathrm{6A}\:\mathrm{and} \\ $$$$\mathrm{ammeter}\:\mathrm{A}_{\mathrm{3}} \:\mathrm{reads}\:\mathrm{0}.\mathrm{4}\:\mathrm{A}\:. \\ $$$$\mathrm{Find}\:\mathrm{reading}\:\mathrm{of}\:\mathrm{ammeter}\:\mathrm{A}_{\mathrm{1}} \:? \\ $$

Answered by ajfour last updated on 07/Aug/18

let V_(ac) =V_0 sin ωt −L(di_2 /dt)=△V_L = −V_(ac) ⇒ i_2 =(V_0 /(ωL))cos ωt ( ∫di_2 =−∫(V_(ac) /L)dt = −∫(V_0 /L)sin ωtdt ) ((∫i_3 dt)/C)=△V_C = −V_(ac) ⇒ i_3 = −ωCV_0 cos ωt i_1 =i_2 +i_3 =((V_0 /(ωL))−ωCV_0 )cos ωt (i_1 )_(rms) =(((V_0 /(ωL))−ωCV_0 )/(√2)) = (((i_2 )_(max) −(i_3 )_(max) )/(√2)) = ((1.6(√2)−0.4(√2))/(√2)) = 1.2 A .

$${let}\:{V}_{{ac}} ={V}_{\mathrm{0}} \mathrm{sin}\:\omega{t} \\ $$$$−{L}\frac{{di}_{\mathrm{2}} }{{dt}}=\bigtriangleup{V}_{{L}} =\:−{V}_{{ac}} \:\:\Rightarrow\:\:{i}_{\mathrm{2}} =\frac{{V}_{\mathrm{0}} }{\omega{L}}\mathrm{cos}\:\omega{t} \\ $$$$\left(\:\int{di}_{\mathrm{2}} =−\int\frac{{V}_{{ac}} }{{L}}{dt}\:=\:−\int\frac{{V}_{\mathrm{0}} }{{L}}\mathrm{sin}\:\omega{tdt}\:\right) \\ $$$$\frac{\int{i}_{\mathrm{3}} {dt}}{{C}}=\bigtriangleup{V}_{{C}} =\:−{V}_{{ac}} \:\:\Rightarrow\:\:{i}_{\mathrm{3}} \:=\:−\omega{CV}_{\mathrm{0}} \mathrm{cos}\:\omega{t} \\ $$$${i}_{\mathrm{1}} ={i}_{\mathrm{2}} +{i}_{\mathrm{3}} \\ $$$$\:\:\:=\left(\frac{{V}_{\mathrm{0}} }{\omega{L}}−\omega{CV}_{\mathrm{0}} \right)\mathrm{cos}\:\omega{t} \\ $$$$\left({i}_{\mathrm{1}} \right)_{{rms}} =\frac{\frac{{V}_{\mathrm{0}} }{\omega{L}}−\omega{CV}_{\mathrm{0}} }{\sqrt{\mathrm{2}}}\:=\:\frac{\left({i}_{\mathrm{2}} \right)_{{max}} −\left({i}_{\mathrm{3}} \right)_{{max}} }{\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\: \\ $$$$\:\:=\:\frac{\mathrm{1}.\mathrm{6}\sqrt{\mathrm{2}}−\mathrm{0}.\mathrm{4}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}}\:=\:\mathrm{1}.\mathrm{2}\:{A}\:. \\ $$$$\: \\ $$

Commented by rahul 19 last updated on 07/Aug/18

ΔV_L =ΔV_C = −V_(AC) ? pls explain ?

$$\Delta\mathrm{V}_{\mathrm{L}} =\Delta\mathrm{V}_{\mathrm{C}} =\:−\mathrm{V}_{\mathrm{AC}} \:?\:\mathrm{pls}\:\mathrm{explain}\:? \\ $$

Commented by ajfour last updated on 07/Aug/18

KVL

$${KVL}\: \\ $$

Commented by ajfour last updated on 07/Aug/18

ammeter readings are the rms values i_(rms) = (i_(max) /(√2)) .

$${ammeter}\:{readings}\:{are}\:{the}\:{rms} \\ $$$${values}\:\:\:{i}_{{rms}} =\:\frac{{i}_{{max}} }{\sqrt{\mathrm{2}}}\:. \\ $$

Commented by rahul 19 last updated on 07/Aug/18

Thank you sir!

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