Test: Electrical Circuits- 3 - Electrical Engineering (EE) MCQ

Given that, the length of a wire having uniform resistance R is stretched in m times with volume remaining same.

Let the length of the wire is L₁

and the area of the wire is A₁

For volume remains to be same

L₁A₁ = L₂A₂

⇒ L₁A­₁ = (mL₁) (A₂)

⇒ A₁ = mA₂

Given that, P = 40 W, V = 110 V

Voltage across resistor = V_s - V = 230 - 110 = 120 V

Power rating of lamp = 40

⇒ VI = 40

We know that, R = R₀[1+α(T₁−T₂)]

Where α is a temperature coefficient of resistance

If the temperature coefficient of resistance is positive, it indicates that the resistance increases with the increase in temperature

If the temperature coefficient of resistance is negative, it indicates that the resistance decreases with the increase in temperature

Watt is the unit of power. We can represent the power as follows

Hence Volt/Ampere is not same as watt, it is equal to ohm.

Total electrons drift across section power second = 10²⁰

We know that, q = it

For t = 1 sec

⇒ q = i

⇒ i = q = ne = 10²⁰ × 1.6 × 10^-19 = 16 A

If one of the resistors gest shorted, remaining two resistors also gets shorted as they are connected in parallel. Hence the current flow through each branch will be zero.

Now voltage across terminal a - b is

The Thevenin’s resistance can be find by neglecting the voltages source.

R_ab = 25 Ω

New Thevenin equivalent network is,

For maximum power transfer, R_L = R_th = 100 Ω

Maximumpower

Norton’s network for any given network can be represented as

The value of the current source is the short circuit current between the two terminals of the network

Norton’s Resistance is the equivalent resistance measured between the terminals of the network with all the energy. Sources are replaced by their internal resistance.

Voltage across resistor R = 15 V

Current(I) = 15/5 = 3A

Power absorbed by resistor =(15)²/5 = 45W

Given that, I₀ = 7.5 mA

Hence, I₁ = 7.5 mA

I₂ = I₀ + I₁ = 15 mA

Now we can reduce the diagram,

As, I₂ = 15 mA, I₃ = 15 mA

I₄ = I₂ + I₃ = 30 mA

Now, the circuit becomes

⇒ V_s = I₄ (R_eq) = 30 × 10^-3 × (8 + 3.5 + 12) = 705 mV

An Electrical network is an interconnection of various electrical elements.

An electrical circuit is a network which has a closed path that gives the return path for the flow of current.

Electrical networks can either be closed or open.

Hence every electrical circuit is a network, but all networks are not circuits.

At steady state, the capacitor acts as open circuit.

At steady state inductor acts as short circuit and the capacitor acts as open circuit.

Given that, V_m = 100 V

V = 86.6 V

f = 50 ⇒ ω = 2πf = 100 π

V = V_m sin ωt

⇒ 86.6 = 100 sin (100 πt)

⇒ sin (100 πt) = 0.866

Given that, I = 5A, V_R = 16 V, V_L = 12 V

V(t) = V_m sin (ωt + θ)

From the given wave form,

V_m = 45

Time period, T = 80 ms

ω = 2πf = 25 π

V(t) = 45 sin (25 πt + 0)

Power factor of first circuit = 0.8 (cos ϕ₁)

Power factor of second circuit = 0.6 (cos ϕ₂)

The power factor of parallel combination,

1) The bandwidth of a parallel RLC circuit is decreases by increasing the resistance

2) The bandwidth is independent of inductance. Hence the bandwidth of the circuit remains the same, if L is increased

3) At resonance the impedance is equal to resistance, it behaves as a pure resistive circuit. Hence the input impedance is a real quantity

4) At resonance, the impedance is equal to resistance. It is the maximum value of impedance and hence current attains its minimum