Electrical Engineering Assignment Sample

Problem 1 Draw an equivalent circuit diagram for the following circuit with minimum number of resistors and capacitors. Calculate value of the resistors and capacitors. (3 Marks)

Solution

Step 1 (Starting from right of the circuit)

• Capacitor  and are in parallel, hence both will replaced by single capacitor of

• Capacitor  and are in parallel, hence both will replaced by single capacitor of.

• Capacitor  and are in parallel, hence both will replaced by single capacitor of.
• Similarly resistance 2.2kΩ and 100Ω are in series, hence they can be replaced by a single resistor of  2.3kΩ, Now we redraw the circuit we get the following circuit as shown below. 

Step 2

• and capacitor are in series and can be replaced by a single capacitor of .

• and capacitor are in series and can be replaced by a single capacitor of.  

Step 3

Further solving the diagram in step 2, we have two capacitors in parallel that can be replaced by single capacitor of we cannot solve the resistor further.

Hence minimum components in circuit are three, two resistor  and one capacitor as show in diagram.

Problem 2 Calculate the reactive power in an industrial power system that power factor and active power are 0.85KW and 28KW respectively

Solution

Given …..…… (i)

…………….. (ii)

As we know that Reactive power =, putting the value from (i) and (ii) we get

Reactive power==

So reactive power in circuit 17.34KVR

Problem 3

Explain what energy flows and where it goes to in a circuit consisting of a battery, resistor, switch and inductor (all connected in series) following the switch’s closure. You will need the formula relating voltage drop across a resistor to current through the resistor and the formula relating voltage across an inductor to the current in the inductor (hint: V=L di/dt) and to get full marks you will have derived an equation for the energy stored in the inductor. A graphical representation of the variation of the energy in the inductor with time is also required.

Solution:

Figure Circuit diagram for RL circuit

Applying Kirchhoff’s voltage law (KVL), we get

As per Ohm’s Law voltage drop across the resistor, R is IR (Ohms Law).

Voltage drop across inductor is given by  

Final expression for voltage drop across the circuit is given by 

From above expression we can see that the voltage drop across the resistor depends upon the current while voltage drop across inductor depends on rate of change of current through the inductor.

Energy Stored in Inductor:

Due to self−inductance opposes the growth of the current. Some work has to be done by external agencies in maintaining the current. Let e is the induced emf then,

Let dw amount is done in dt time then , putting the value of e we get

For total work done we need to integrate from 0 to

Negative sign represents the oppose of cause, hence energy stored in inductor is given by  

Figure charging and discharging of inductor