Network Elements - 1 - Free MCQ Practice Test with solutions, GATE EE Theory

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

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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

B is correct; the current supplied by the battery is 0.5 A.

Let I1 be the current through the left 1 Ω resistor (from battery to node A) and I2 be the current flowing through the top-right 1 Ω resistor (and also through the rightmost 1 Ω resistor). The dependent current source injects a downward current equal to I2 at node A.

Applying KCL at node A: the incoming current I1 equals the sum of currents leaving to the right and downward, so I1 = 2 I2.

Applying KVL around the outer loop (battery and the three 1 Ω drops): the supply 1 V equals the sum of voltage drops I1×1 Ω + I2×1 Ω + I2×1 Ω, giving I1 + 2 I2 = 1.

Substitute I1 = 2 I2 into I1 + 2 I2 = 1: 2 I2 + 2 I2 = 1 ⇒ 4 I2 = 1 ⇒ I2 = 1/4 A.

Then I1 = 2 I2 = 2 × 1/4 = 1/2 A = 0.5 A. Therefore the battery current is 0.5 A, corresponding to option B.

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.