Question

Don't answer directly..

hey plz Solve BOTH questions​

Answer 1

Answer:

Question 1:

after steady state is achieved, the capacitor branch is disconnected

net resistance = 2.8 + 3×2/(3+2) = 2.8 + 1.2 = 4.0ohms

current through battery = 6/4 = 1.5A

Voltage across 2 ohm resistor = 6 - 2.8×1.5 = 1.8V

So current through 2ohm resistor

= 1.8/2 = 0.9A

Answer is (a)

Question 2:

Let's calculate current first..

5 - 10 - i×1 - i×1.5 = 0

| i | = 5/2.5 = 2A

So,

∆V_AB = 10 - 2×1 = 8V

Answer 2

Answer:

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