SSC JE 2014 Electrical question paper with solution(Morning-Shift)

Let us assume the capacitor was 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 the resistor and voltage drop across the capacitor is equal to the battery voltage. Hence at t = ∞ the capacitor acts as an open circuit.

Danger 440V” plates are caution notice.

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

There are mainly two functions of the choke coil

1.  Limit the current.
2. Produce high voltage across tube light.

Limit the current

• In a gas discharge, such as a fluorescent lamp, current causes resistance to decrease. This is because as more electrons and ions flow through a particular area, they bump into more atoms, which frees up electrons, creating more charged particles.
• In this way, the current will climb on its own in a gas discharge, as long as there is adequate voltage (and household AC current has a lot of voltage).
• If the current in a fluorescent light isn’t controlled, it can blow out the various electrical components.

Produces High voltage Across Tubelight

• Choke is nothing but the coil/ballast ( inductor) which is used to induce the high voltage across it. as we know that inductor has the property to induce high voltage for a brief period of time, this high voltage is required to ionize the gases in the starter.

synchronous speed

Ns =120f/P = 120 x 50/4 = 1500 RPM

% S = (Ns – Nr) x 100/Ns

4 =(1500 – Nr) x100/1500

60 = 1500 – Nr

Nr = 1500 – 60 = 1440 RPM

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

Hysteresis motor doesn’t have teeth or winding, therefore, these motor are free from mechanical vibration

In case of series.

When motors are connected in series and are in running position, Speed ∝ (voltage) ∝(V/2) (i) (Since the voltage across each motor = V/2 ) and current is Ia.

Ns ∝ V/2I

In case of parallel

When the motor is connected in parallel and are in running position, Speed ∝ (voltage) (V). (Since voltage across each motor = V ) and current is I_a/2

Np ∝ 2V/I

From both the equation

Ns/Np = V/2I/2V/I

N parallel = 4Nsereis

According to Millman’s theorem

Eeq = ∑EiYi/∑Yi

I1 = 8/3

I2 =16/3

I3 = 24/3 = 8A

Eeq = (8/3 + 16/3 + 8)/ (1/3 + 1/3 + 1/3) =16 V

Zeq = 1/∑Yi

= 1/(1/3 + 1/3 + 1/3) = 1Ω

Now the circuit become

I = 16/4 = 4 A

An ideal voltage source is a voltage source having no internal resistance.

This means there is no voltage drop inside the source, so one gets the same voltage level at the terminals of the source as produced inside the source.