Test: Electrical Circuits- 2 - Electrical Engineering (EE) MCQ

The internal resistance of an ideal voltage source should be zero and practically it is very low.

Given that, source resistance is 1 ohm

Hence the source will be a voltage source

Rating of lamp = 60V, 40 W

When lamp is connected, maximum current that can be flow through the circuit is 

Let R_S be the series resistance.

By KCL,

The sum of current entering a junction is equal to the sum of current leaving at that junction.

⇒ Sum of currents leaving = 9 A

This 9 A current will flow through three resistances having same value.

The current will be some through each resistor

Given that, voltage source supplies 24 W

By applying KVL,

- 24 + I₁ R₁ + I₂ R₂ = 0

⇒ I₂ R₂ = 24 – R₁ → (1)

Given that, current source supplies 6 W

By KVL in second loop

I₂ R₂ = 6 → (2)                

From (1) and (2)

⇒ R₁ = 18 Ω

By KCL at point ‘O’

⇒ I₂ = 1 + I₁ = 2 A

We know that, I₂ R₂ = 6

⇒ R₂ = 3 Ω

By KCL at point ‘O’

I₁ = I₂ + 2 = 0.5 + 2 = 2.5

To find Norton’s resistance, we need to short circuit the voltage source and open circuit the current source

R_N = (4 + 4) || 8 = 4 Ω

To find the Norton’s current, we need to find short circuit current across terminals a-b

By converting Delta (ABC) in to star

R_eq = 18 || (14 + 4) = 9 Ω

Let the current passing through 6 Ω resistor is I.

By source transformation,

By source transformation,

By current division, 

To find Norton’s current, we need to find the short circuit current across the terminals a and b.

By source transformation,

By KVL,

-180 + 20 I_SC + 120 +40 I_SC = 0

⇒ I_SC = 1 A

According to maximum power transfer theorem, in AC circuits the load impedance should be complex conjugate of the internal impedance of the active network

C_eq = 10 + 20 + 40 = 70 μF

Charge, Q = CV = (70 μ F) (100 V) = 7 × 10^-3 C

V(t) = 10 cos (400 π t)

ω = 400 π

⇒ 2 π f = 400 π

⇒ f = 200 Hz

V_m = 10 V

Given that, inductance = 0.2 H

Voltage = 200 V

Inductive reactance = 2 π fL = 2 π × 50 × 0.2 = 62.8 Ω

Given that, R = 2 Ω, L = 1 mH, C = 0.1 μ F

For series resonant circuit,

Maximum value of V(t) = 2 V

Given that, V = 50 sin (ω t + 60°)

 i = 10 cos (ω t)

i = 10 cos (ω t)

i = 10 sin (ω t + 90°)

i is leading V

Phase difference (ϕ) = 30°

Z = 8 +j 6

Power consumed by the load

= 4229.11 W

Given that, X_L = 1000 Ω

X_C = 1000 Ω

R = 0.1 Ω

F₀ = 10 MHz