Step 1:

The bridge network is shown below.

Apply KVL for the loop ABDA.

Apply KVL for the loop BCDB.

Apply KVL for the loop ADCEA.

Step 2:

Now solve the three equations.

Multiply equation (1) with 2.

Subtract equation (2) from equation (4).

Multiply equation (2) with 18.

Multiply equation (3) with 12.

Add equation (6) and equation (4).

Substitute in the above equation.

The current flowing through 5 ohms resistor is .

Solution:

The current flowing through 5 ohms resistor is .

(2.1)

Step 1:

The resistance of the circuit is .

The inductance of the circuit is 0.2 H.

The capacitance of the circiut is .

The applied voltage is 100 V.

Input frequency is 50 Hz.

The elements are connected in series, hence equilvalent resistance .

Impedance due to resistor element is .

Impedance due to inductor element is

Impedance due to capacitive element is

Equivalent inpedance : .

Solution :

Total impedance is 

(2.2)

Step 1:

Find the total current.

Applied voltage is 100 v.

Total impedance of the circuit is      (From 2.2)

Total current in the circuit is .

Total current in the circuit is .

Solution :

Total current in the circuit is .

(2.3)

Step 1:

Find the voltage drop across the resistor.

Current across each element is same as they are connected in series.

Voltage drop across resistor : .

Voltage drop across the resistor : .

Solution :

Voltage drop across the resistor : .

(2.4)

Step 1:

Find the voltage drop across the inductor.

Current across each element is same as they are connected in series.

Voltage drop across resistor : .

Impedance due to inductor element is .

Voltage drop across inductor : .

Solution :

Voltage drop across inductor : .