Test: Basic Concepts of Network Theory - Electrical Engineering (EE) MCQ

From Ohm’s law,

V₂= -3i and V₁=2i

Appling the KVL, -20 + V₁ – V₂=0 ———–(i)

Put the value on above equation and get i = 4A

Hence, V₂ = -3 x 4 = -12 V

Diode is non-linear elements is removed.

After shorting the voltage source the redrawn circuit is

Thevenin resistance

= 11/3 ohm

• Based on kirchoffs current law, we know that incoming currents at a node is equals to outgoing currents.
• So, based on this 4 and 3 will get canceled with 5 and 2.
• So, finally i = 1.

• In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise. 
• Applying KVL we get:
  v₁ + 60 - 100 = 10 x 20
  v₁ = 240 V

Going from 10 V to 0 V:

• 10 + 5 + E + 1 = 0
• E = -16V
• Minus sign indicates that the polarity of battery should be reversed.

• 100 = 65 + v₂
  ⇒ v₂ = 35 V

• v₃ - 30 = v₂ 
  ⇒ v₃ = 65 V
• 105 + v₄ - v₃ - 65 = 0
  ⇒ v₄ = 25 V
• v₄ + 15 - 55 + v₁ = 0
  ⇒ v₁ = 15 V 

► V = W / Q
► Q = W / V = 660 / 110 = 6 C

Total voltage=Voltage drop across 60 Ω resistor+Voltage drop across R1​+Voltage drop across R2​

The current i will be distributed in the cube branches symmetrically

If we go from +ve side of 1 kΩ through 7 V, 6 V and 5 V
We get, v₁ = 7 + 6 - 5 = 8 V

Let's assume the current source as the ideal source:
Apply KVL in the loop, we get
v - 4 × 2.5 - 10 = 0
⇒ v = 20 V

• Cosɸ = R/Z; where R= 10 ohms and Z = 20 ohms  
• Cosɸ = 1/2 means phase angle is 60 degree.

We know that,
I=Q / t or Q = I x t = 10 x 1 hours
{since unit of current is Cs^-1, therefore time should be in seconds}
∴ Q = 10 x 60 x 60
= 36000 C
= 3.6 x 10⁴ C

12 * C  = 2 m x 10  
C = 1.67 mF
Hence, Capacitance is equal to 1.67 mF.

This 0.2 V increases linearly from 0 to 0.2 V. Then current is zero. So capacitor hold this voltage.

 

= 600 mA
For 1 ms < t < 2 ms, 

= 3125 t²
At t = 4 ms, v_c = 0.05 V
It will be parabolic path at t = 0
t-axis will be tangent.