Test: Network Elements - 1 - Electrical Engineering (EE) MCQ

Sign convention for Power delivered and absorbed:
When current is leaving the positive voltage terminal of the element or current is entering the negative voltage terminal of the element then the particular element is delivering power.
∴ P = +Vi

• When current is entering the positive voltage terminal of the element or current is leaving the negative voltage terminal of the element then the particular element is absorbing power.

∴ P = -Vi

• The power delivered to an element is denoted by the positive sign, whereas, power supplied by an element is denoted by the negative sign.
• In figure (i), when current enters through the positive terminal of the circuit, the power has a positive sign.
• In figure (ii) when the current leaves the positive terminal, the power has a negative sign

Calculation:
Given that, 
V = 2 volts, 
i = 3 A

In the above circuit, the current is leaving the positive voltage terminal of the element or the current is entering the negative voltage terminal of the element then the particular element is delivering power.
Thus, the power drawn by the circuit shown in the above figure can be calculated as
P = + Vi
=> P = +(2 × 3)
∴ P = + 6 W

Reraw the circuit:

Now, using voltage division,

When resistances are connected in parallel, the equivalent resistance is given by

When resistances are connected in series, the equivalent resistance is given by

Calculation:
Given that R₁ = R₂ = R₃ = 6 Ω and all are connected in parallel

​​​​​​​

Electric current: If the electric charge flows through a conductor, we say that there is an electric current in the conductor.
If Q charge flow through the conductor for ‘t’ seconds, then the current given by that conductor is

Q = I × t
I = current
t = times

Calculation:
Given I = 5 amp
t = 3 min = 180 sec
Q = I × t
Q = 5 × 180 = 900 C

Using KCL at node , we get
I + 0.4I = 14
I= 10A
Now, Power supplied,
P = 25 x 10  = 250W

So, V_out = (5 x 2) + 10 = 20V

Applying KCL at node A,
-I₁ + I₂ + I₂ = 0
2I₂ = I₁  .... (i)
and applying KVL in loop ABCD,
1 - I₁ - I₂ -  I₂ = 0
I₁ + 2I_{2  = 1 .... (ii)}
From equation (i) and (ii),

When two or more electrical devices of the same voltage are connected in parallel connection, their equivalent power rating can be calculated as
P_eq = P₁ + P₂ + P₃ + ……
When ‘n’ number of devices of the same voltage and same power rating (P) are connected in parallel connection, their equivalent power rating can be calculated as
P_eq = nP
When two or more electrical devices of the same voltage rating are connected in a series connection, their equivalent power rating can be calculated as

When ‘n’ number of devices of the same voltage and same power rating (P) are connected in series connection, their equivalent power rating can be calculated as

Calculation:
When two identical bulbs are resistors are connected in series across a battery, their equivalent power (Peq) is ‘10 W’

∴ The power rating of each bulb, P = 20 W

When these two resistors are connected in parallel, then the equivalent power dissipated is
P_eq = 2 × P
P_eq = 40 W

Alternate Approach:
Series circuit: 
In the series circuit, the voltage drop across each resistor = V/2
The total current through the circuit = V/2R amps.
The power dissipated by each resistor = V/2 × V/2R = V²/4R
Which must be equal to 5 Watts in order to make a total of 10 W.

Parallel circuit:
The voltage drop across each resistor = V.
The current through each resistor = V/R.
Multiplying these gives V2/R which is 4 times the series value of V²/4R
∴ The total power dissipated in parallel circuit = 4 × 10 = 40 W

Linear Element: An element is linear if :

• It follows additivity and superposition
• V-I graph is a straight line passing through the origin

Time variant Element: An element is said to be time-variant if the V-I graph varies with time.

Application:
Given V-I characteristic : v(t) = i2(t)
The characteristics can be drawn as,

It is not a straight line so non-linear
It is time-invariant because in any network analysis problem we take the circuit parameters V and I to be time-invariant.

Alternate Method:

Shifting the input, we get:
v₁(t) = i₁² (t - t1)
Now, shifting the output with the same amount, we can write:
v₂ (t) = v₁(t - t₁) = i₁² (t - t₁) = v₁ (t)

Since v₁(t) = v₂(t), the system is time invariant.