SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-2)

Answer.4. 2.05 μF

Explanation:-

Total capacitance between point L and N

=2 + 1 = 3μF

Now total capacitance between L and O

= 1 + (3 x 2)/(3 + 2) = 2.2μF

Total capacitance Between L and M

= 1 + (2.2 x 2)/(2.2 + 2) = 2.05μF.

Answer.3. 0.1s

Explanation:-

Time constant of RL series circuit is given by

τ = L/R

τ = 2/20 = 0.1 sec

Answer.3. 40 V

Explanation:-

I = V/R = 120/(100 + 50) = 120/150 = 4/5 A

V_CB = 50 x 4/5 = 40 V

Now voltage at ‘c’ w.r.t to ground = Vc

V_CB = Vc – Vb

Vc = V_CB – V_B

= 40 + 0 = 40 V

Answer.1. 3R Ω

Explanation:-

The value of equivalent delta connected resistor

R1 + R2 + R1R2/R3

since all the resistor are of RΩ

R + R + R²/R = 3RΩ

Answer.4. Linear Bilateral Networks

Explanation:-

The superposition theorem for electrical circuits states that for a linear system the response(voltage or current) in any branch of a bilateral linear circuit having more than one independent source equals the algebraic sum of the responses caused by each independent source acting alone, where all the other independent sources are replaced by their internal impedances

Answer.3. 50 Hz

Explanation:-

Magnetic energy stored in an inductor = LI²/2.

The energy stored is maximum when the current is maximum and energy stored is minimum when the current is minimum (i.e. zero). ln, an a.c. circuit, current goes from the maximum value to zero in T/4 sec. where T is the time period of a.c.

T/4 = 5

T = 20 ms = 20 x 10^-3

Frequency  = 1/T = 1/20 x 10^-3 = 50 Hz

Answer.4. 60 W lamp filament has a longer length

Explanation:-

Let’s suppose the voltage is the same for both the bulb

Power P = V²/R

P = V²A/ρl………………. (since specific resistance  R= ρ (L/A).

L =V²A/ρP

Since both, the filament has the same thickness. So ‘A’ voltage will be same

l ∝ 1/P

Hence Filament of low power will have a longer length.

Note:- If P ↑ R↓ L ↓

For more watt bulb, less resistance will be there so, the length of that bulb will be lower.

Answer.1. Open-circuit

Explanation:-

Let us assume the capacitor initially uncharged with the switch opened.

After the switch is closed, the battery begins to charge the plates of the capacitor and the charge passes through a resistor. As the capacitor is being charged, the circuit carries a changing current. The charging process continues until the capacitor is charged to its maximum equilibrium value, Q = CV, where V is the maximum voltage across the capacitor Once the capacitor is fully charged, the current in the circuit is zero.

The current approaches 0 at t = ∞ because there is no voltage drop across resistor and voltage drop across the capacitor is equal to the battery voltage. Hence at t = ∞ the capacitor acts as an open circuit.

Answer.4. 25 Kilo ohm

Explanation:-

For the given circuit, the value of current passing through R3 = (50 -10) =40 mA

R₂ And R₃ are in parallel, therefore, voltage across them will be same

R₃ x 40 x 10^-3 = R₂ x 10 x 10^-3

R3 = (100 x 10³ x10 x 10^-3)/(40 x 10^-3)

25 x 10³ = 25 kΩ

Answer.1. ± 6%

According to Indian Electricity Rules, the permissible voltage drop from the supply terminal to any point on the wiring system should not exceed ±6% volt of the nominal supply voltage.