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

Ques 1. Two 10 ohm resistance is connected in parallel, their effective resistance will be (SSC-2017)

5

10

7

None of the above

Answer.1. 5 ohm

Explanation:

Since the resistance are connected in parallel, therefore, their effective resistance will be

R_{1} x R_{2} / R_{1} + R_{2}

100/20 = 5 ohm

Ques.2. Which efficiency is more in case of the cell (SSC-2017)

Watt-hour efficiency

Ampere-hour Efficiency

Output Power efficiency

None of these

Answer.2. Ampere-hour Efficiency

Explanation:-

Actually battery capacity is measured in the amount of charge it can hold. Now we know the current is charge per unit time, therefore the product of current and time gives us the charge. That is why battery capacities are measured in milliampere-hour in case of cellphone batteries or ampere-hour in case of automotive batteries.

Battery capacity = charge stored = current x time or amp-hr.

The efficiency of a cell can be considered in two ways:

The quantity or ampere-hour (Ah) efficiency

The energy or watt-hour (Wh) efficiency

The Ah efficiency does not take into account the varying voltages of charge and discharge. The Wh efficiency does so and is always less than Ah efficiency because average p.d. during discharging is less than that during charging. Usually, during discharge the e.m.f. falls from about 2.1 V to 1.8 V whereas during charge it rises from 1.8 volts to about 2.5 V.

Ques.3. Superposition theorem is based on the concept of the opposite of which is (SSC-2017)

Linearity

Causality

Non-Linearity

Instability

Answer.3. Non-Linearity

Explanation:

Superposition theorem is based on circuit linearity property and the opposite of Linearity is Non-Linearity.

Ques.4. Who Discovered the neutron? (SSC-2017)

James Chadwick

Rutherford

Nikola Tesla

Einstein

Answer.1. James Chadwick

Explanation:

James Chadwick discovered the neutron in 1932.

by Ques.5. The phase difference between voltage and current wave through a circuit element is given as 30. The essential condition is that (SSC-2017)

Both waves must have the same frequency

Both waves must have identical peak values

Both waves must have zero value at the same time

None of the above

Answer.1. Both waves must have the same frequency

Explanation:

Consider the given circuit

The current would be 0 if the voltage of both the sources is the same and the AC sources are in phase i.e at the same frequency.

However, if the same frequency sources are connected together but if they have phase difference even then current will flow.

If you connect voltage sources with the different phase, voltage, frequency then the high magnitude of current will flow in the circuit. This can damage the sources. This can cause metallic wires to melt because of heating.

Hence if both waves have the same frequency than the phase difference between voltage and the current wave is 30°.

Ques.6. In the series resonant circuit, the impedance of the circuit is (SSC-2017)

Minimum

Maximum

Zero

None of these

Answer.1. Minimum

Explanation:

The total impedance of the series LCR circuit is given as

Z = R + j (X_{1} – X_{²})

where X_{1} is inductive reactance

and X_{2} is capacitive reactance.

At a particular frequency (resonant frequency), we find that X_{1}=X_{2 }because the resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase.

Ques.7. In R-L-C series resonant circuit magnitude of resonance frequency can be changed by changing the value of (SSC-2017)

R only

L only

C only

L or C

Answer.4. L or C

Explanation: An electric circuit generally consists of circuit elements like resistance, inductance, and capacitance. The voltage and frequency are generally constant at the supply terminal. However, in electronic communication circuits, the supply voltage may have a variable frequency. When the frequency is variable, the inductive and capacitive reactance of the circuit element will change. (X_{L} = 2.π.f.L and X_{C} = 1 ⁄ 2.π.f.C) Hence by changing the value of either capacitor or inductor in the series resonant circuit then the frequency of the circuit will change accordingly

Ques.8. If the length of the material remains the same while the diameter of material becomes half, what will be the new resistance (SSC-2017)

Four times

Doubles

Half

No change

Answer.1. Four Times

Explanation:

Resistance = R= pL/A

Where p is the resistivity of the material

L is length

A is a cross-section area

A =πr^{2}

Therefore if we reduce the diameter by 1/2, i.e { A = Π (D/2)^{2}} then the resistance becomes 4 times

Ques.9. The time constant of the series RC circuit is given by (SSC-2017)

R/C

RC^{2}

RC

R^{2}C

Answer.3. RC

Explanation:-

Consider the series RC circuit. Considered the capacitor is 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 the resistor. As the capacitor is being charged, the circuit carries a changing current. Because of the presence of the resistor the capacitor is not charged all at once but gradually over time. The charging process continues until the capacitor is charged to its maximum equilibrium value, Q = Cε, where ε is the maximum voltage across the capacitor.

Once the capacitor is fully charged, the current in the circuit is zero.If we assume the capacitor is uncharged before the switch is closed, and if the switch is closed at t = 0, we find that the charge on the capacitor varies with time according to the equation

q = Q(1– e^{-t/RC})

The RC time constant, also called tau, the time constant (in seconds) of an RC circuit, is equal to the product of the circuit resistance (in ohms) and the circuit capacitance (in farads), i.e.

τ =RC

The time constant represents the time required for the charge to increase from zero to 63.2% of its maximum equilibrium value. This means that in a period of time equal to one time constant, the charge on the capacitor increases from zero to 0.632Q.

Ques.10. Which of the following is not a unit of inductance (SSC-2017)

Henry

Columb/Volt ampere

Volt second per ampere

All of the above

Answer.3. Volt second per ampere

Explanation:

The unit of measurement for the inductance of an inductor is the henry, which is equal to 1 volt-second per ampere.

The voltage across an inductor is proportional to the rate of change of current (measured in amps/second) through it. The larger the inductance, the larger the voltage across the inductor for the same rate of change of current. If the current is flowing left to right through an inductor and increasing (respectively decreasing) with time, then the potential on the left side of the inductor is higher (respectively lower) than the potential on the right side.