COMEDK Exam 2019 Solved Paper

In the steady state analysis capacitor acts as an open circuit. Therefore, the equivalent circuit is given asThus, current in the 4Ω resistor is zero. 2Ω and 3Ω resistors are parallel to each other. Thus, the resistance in the upper branch is given by
R=‌

2×3

2+3

=‌

6

5

Ω
The upper branch resistor are connected to 2.8Ω resistor in series. Therefore, equivalent resistance in the circuit is given by
Req‌=28+‌

6

5

‌=‌

14.0+6

5

=4Ω
Current, I=‌

‌ Voltage ‌

‌ Resistance ‌

=‌

6

4

=15A
Current in the 2Ω resistor is given by current division rule
i.e. ‌‌I2Ω=I×‌

3Ω

2Ω+3Ω

⇒‌‌I2Ω=1.5×‌

3

5

=0.9A