The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q₀ e^ -R t 2 L , where Q₀ is the charge at t=0 ~s . The time at which charge amplitude decreases to 0.50 Q₀ is nearly: [Given that R=1.5 , L=12 mH , (2)=0.693

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The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q₀ e^ -R t 2 L , where Q₀ is the charge at t=0 ~s . The time at which charge amplitude decreases to 0.50 Q₀ is nearly: [Given that R=1.5 , L=12 mH , (2)=0.693 ]

Quantrex Academy · Accepted Answer

Correct answer: B. 11.09 ms

Given Q=Q₀ e^ -R t / 2 L R=1.5 , L=12 mH , In (2)=0.693Q=0.5 Q₀, t= ? 0.5 Q₀=Q₀ e^ -R t / 2 L 1 2 =e^ -R t / 2 L Taking log on both sides ( 1 2 )= e^ -R t / 2 L 2= R t 2 L t= 2 L In 2 R = 2 12 10⁻³ 0.693 1.5 t=11.09 ~ms

The amplitude of the charge oscillating in a circuit decreases exponentially as Q=Q₀ e^ -R t 2 L , where Q₀ is the charge at t=0 ~s . The time at which charge amplitude decreases to 0.50 Q₀ is nearly: [Given that R=1.5 , L=12 mH , (2)=0.693 ]

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