આકૃતિ Q46માં દર્શાવેલો પરિપથ ધ્યાનમાં લો. સ્વિચ લાંબા સમયથી બંધ છે અને t = 0s પર ખૂલે છે. જો t > 0s માટે પ્રવાહ i(t) = 2.5 + 1.5e^(–4t) A હોય, તો R1 અને R2 નું મૂલ્ય શું છે?
જવાબ અને પગલાવાર ઉકેલ
A — R1 = 3 Ω, R2 = 5 Ω. RL પ્રતિભાવ i(t) = I∞ + (I₀ − I∞)e^(−t/τ) થી I∞ = 2.5 A, I₀ = 4 A અને τ = 1/4 s. સ્વિચિંગ પહેલાં/પછીના DC સમતુલ્ય અને τ = L/Req લાગુ કરતાં R1 = 3 Ω, R2 = 5 Ω.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept and Reasoning
For an RL switching response i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ, the given expression identifies I∞=2.5 A, the initial current as 4 A, and τ=1/4 s. Applying the pre-switch and post-switch DC equivalents and τ=L/Req gives R1=3 Ω and R2=5 Ω.
Calculation / Key Relation
i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ
τ=0.25 s
R1=3Ω, R2=5Ω
Exam Takeaway
Read I∞, I0+ and τ directly from the exponential form first.
Common Mistake
Do not assume the coefficient of e⁻⁴ᵗ is the final current.
Quick Trick
Read I∞, I0+ and τ directly from the exponential form first.
Consider the circuit shown in Fig. Q46. The switch is initially closed for a long time and opens at t = 0s. What is the value of R1 and R2, if current i(t) = 2.5 + 1.5e– 4t A for t > 0s? Fig. Q46
Answer & step-by-step solution
For an RL switching response i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ, the given expression identifies I∞=2.5 A, the initial current as 4 A, and τ=1/4 s. Applying the pre-switch and post-switch DC equivalents and τ=L/Req gives R1=3 Ω and R2=5 Ω.
Concept and Reasoning
For an RL switching response i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ, the given expression identifies I∞=2.5 A, the initial current as 4 A, and τ=1/4 s. Applying the pre-switch and post-switch DC equivalents and τ=L/Req gives R1=3 Ω and R2=5 Ω.
Calculation / Key Relation
i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ
τ=0.25 s
R1=3Ω, R2=5Ω
Exam Takeaway
Read I∞, I0+ and τ directly from the exponential form first.
Common Mistake
Do not assume the coefficient of e⁻⁴ᵗ is the final current.
Quick Trick
Read I∞, I0+ and τ directly from the exponential form first.
चित्र Q46 में दिखाए गए परिपथ पर विचार कीजिए। स्विच लंबे समय से बंद है और t = 0s पर खुलता है। यदि t > 0s के लिए धारा i(t) = 2.5 + 1.5e^(–4t) A है, तो R1 और R2 का मान क्या है?
उत्तर और चरणबद्ध हल
A — R1 = 3 Ω, R2 = 5 Ω। RL अनुक्रिया i(t) = I∞ + (I₀ − I∞)e^(−t/τ) से I∞ = 2.5 A, I₀ = 4 A और τ = 1/4 s। स्विचिंग से पहले/बाद के DC समतुल्य और τ = L/Req लागू करने पर R1 = 3 Ω, R2 = 5 Ω।
विस्तृत हल (मूल भाषा में)
Concept and Reasoning
For an RL switching response i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ, the given expression identifies I∞=2.5 A, the initial current as 4 A, and τ=1/4 s. Applying the pre-switch and post-switch DC equivalents and τ=L/Req gives R1=3 Ω and R2=5 Ω.
Calculation / Key Relation
i(t)=I∞+(I0+−I∞)e⁻ᵗ/τ
τ=0.25 s
R1=3Ω, R2=5Ω
Exam Takeaway
Read I∞, I0+ and τ directly from the exponential form first.
Common Mistake
Do not assume the coefficient of e⁻⁴ᵗ is the final current.
Quick Trick
Read I∞, I0+ and τ directly from the exponential form first.
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