SSC JE Electrical Questions with Detailed Solutions

​$$\text{Given Values:} \\[4pt]V_{CC} = 10\,\text{V} \\[4pt]R_C = 8\,\text{k}\Omega \\[4pt]I_B = 15\,\mu\text{A} \\[4pt]\beta = 40 \\[8pt]\textbf{Formulas Used:} \\[4pt]\bullet \; \text{Collector current:} \quad I_C = \beta \cdot I_B \\[6pt]\bullet \; \text{Emitter current:} \quad I_E = I_C + I_B \\[6pt]\bullet \; \text{Collector-emitter voltage:} \quad V_{CE} = V_{CC} - I_C \cdot R_C \\[8pt]\text{First, calculate the collector current (} I_C \text{):} \\[4pt]I_C = 40 \times 15 \times 10^{-6} = 0.6\,\text{mA} \\[8pt]\text{Now calculate the collector-emitter voltage (} V_{CE} \text{):} \\[4pt]V_{CE} = 10 - (0.6 \times 8) = 5.2\,\text{V}$$​​

$$\text{To calculate the current, we can use the formula:} \\[6pt]I = \dfrac{P}{V} \\[10pt]\text{Where:} \\[4pt]\bullet P = 2\,\text{kW} = 2000\,\text{W (Power)} \\[4pt]\bullet V = 240\,\text{V (Voltage)} \\[8pt]\text{Now, applying the values:} \\[6pt]I = \dfrac{2000}{240} = 8.33\,\text{A}$$​​

Dependent sources are categorized based on their type of control. There are two types of dependent sources:
Voltage-controlled voltage source (VCVS):

Current-controlled voltage source (CCVS):

Voltage-controlled current source (VCCS):

Current-controlled current source (CCCS):

$$\text{Given:} \\[4pt]\bullet I = \text{Current flowing through the conductor} \\[4pt]\bullet t = \text{Time in seconds} \\[4pt]\bullet R = \text{Resistance of the conductor} \\[8pt]\text{The electrical energy supplied in the heating effect is given by:} \\[6pt]E = I^2 R t \\[8pt]\text{Where:} \\[4pt]E = \text{Electrical energy supplied (in joules)} \\[4pt]I = \text{Current (in amperes)} \\[4pt]R = \text{Resistance (in ohms)} \\[4pt]t = \text{Time (in seconds)}$$​​

To prevent rusting and corrosion in electric irons, the plates of the bottom surface and edges are typically made of heavy chromium. Chromium is highly resistant to rust and corrosion, which is why it is commonly used as a protective coating for metals exposed to moisture and heat. This coating helps to maintain the durability and functionality of the electric iron.
 Mica is used for insulation purposes, not for preventing rusting.
 Asbestos was once used for heat resistance but is not used for rust prevention.
 Heavy iron may rust over time and does not provide the same protection as chromium.

$$V = I \times (R + r) \\[8pt]\text{Where:} \\[4pt]\bullet \; V \text{ is the total voltage} \\[4pt]\bullet \; I \text{ is the current} \\[4pt]\bullet \; R \text{ is the load resistance} \\[4pt]\bullet \; r \text{ is the internal resistance of the battery} \\[8pt]\textbf{Calculation:} \\[4pt]\text{Given:} \\[4pt]\bullet \; V = 20\,\text{V} \\[4pt]\bullet \; R = 19\,\Omega \\[4pt]\bullet \; I = 1\,\text{A} \\[8pt]\text{Using the formula:} \\[4pt]20 = 1 \times (19 + r) \\[4pt]20 = 19 + r \\[4pt]r = 20 - 19 \\[4pt]r = 1\,\Omega$$​​

$$\text{Given:} \\[4pt]\bullet \; \text{Supply Voltage, } V_T = 16\,\text{V} \\[4pt]\bullet \; \text{Forward Voltage of Bulb, } V_{\text{Bulb}} = 4\,\text{V} \\[4pt]\bullet \; \text{Current, } I = 40\,\text{mA} = 0.04\,\text{A} \\[4pt]\bullet \; \text{Voltage drop across the series resistor, } V_R = V_T - V_{\text{Bulb}} \\[4pt]\bullet \; \text{The value of the series resistor is given by } R = \frac{V_R}{I} \\[8pt]\text{Calculate } V_R: \\[4pt]V_R = V_T - V_{\text{Bulb}} = 16\,\text{V} - 4\,\text{V} = 12\,\text{V} \\[8pt]\text{Now calculate the value of the resistor } R: \\[4pt]R = \frac{V_R}{I} = \frac{12\,\text{V}}{0.04\,\text{A}} = 300\,\Omega$$​​

Explanation: A periodic signal is a signal that repeats its pattern at regular intervals, meaning it has a definite pattern that keeps coming back after a specific time period. 
Dependent signal: This term might refer to a signal that depends on another signal's values, but it doesn't necessarily imply a repeating pattern.
Independent signal: An independent signal is one that doesn't depend on any other signal, but it also doesn't necessarily have a repeating pattern. 

To solve this problem, we can use the formula for the energy stored in an inductor:

​$$E=\frac{1}{2}LI^2$$​​

Where:

• E = the energy stored (in joules)
• L = the inductance (in henries)
• I = the current (in amperes)

First, let's convert the given energy from kilowatt-hours (kWh) to joules:

1 kWh=3600 kJ1

​$$4\ \text{kWh}=4\times3600=14400\ \text{kJ}=14400\times10^3\ \text{J}$$​​

Now, substitute the known values into the formula:

$$\text{To solve this problem, we can use the formula for the energy stored in an inductor:} \\[6pt]E = \dfrac{1}{2} L I^2 \\[8pt]\text{Where:} \\[4pt]\bullet E = \text{the energy stored (in joules)} \\[4pt]\bullet L = \text{the inductance (in henries)} \\[4pt]\bullet I = \text{the current (in amperes)} \\[8pt]\text{First, let's convert the given energy from kilowatt-hours (kWh) to joules:} \\[4pt]1\,\text{kWh} = 3600\,\text{kJ} \\[4pt]4\,\text{kWh} = 4 \times 3600 = 14400\,\text{kJ} = 14400 \times 10^3\,\text{J} \\[8pt]\text{Now, substitute the known values into the formula:} \\[6pt]14400 \times 10^3 = \dfrac{1}{2} L (2000)^2 \\[8pt]14400 \times 10^3 = \dfrac{1}{2} L \times 4000000 \\[8pt]L = \dfrac{14400 \times 10^3 \times 2}{4000000} \\[8pt]L = \dfrac{28800 \times 10^3}{4000000} \\[8pt]L = 7.2\,H$$​​

To calculate the maximum current level for a 4Ω resistor with a maximum power dissipation of 784 W, we can use the formula for power dissipation in a resistor:

​$$P = I^2 R \\[8pt]\text{Where:} \\[4pt]\bullet \; P \text{ is the power dissipation (784 W),} \\[4pt]\bullet \; I \text{ is the current (which we need to find),} \\[4pt]\bullet \; R \text{ is the resistance (4}\Omega\text{).} \\[8pt]\text{Rearranging the formula to solve for } I: \\[4pt]I = \sqrt{\frac{P}{R}} \\[8pt]\text{Now, substituting the known values:} \\[4pt]I = \sqrt{\frac{784}{4}} = \sqrt{196} = 14\,A \\[8pt]\text{Thus, the maximum current is } 14\,A.$$​​

Resistance is a property that is independent of frequency in direct current circuits. This means that the frequency of the current flowing through a resistor does not influence the value of its resistance. Therefore, when selecting a resistor, the frequency range is not a relevant consideration.

The maximum possible mutual inductance M between two inductively coupled coils can be calculated using the formula:

$$\\[6pt]M = \sqrt{L_1 L_2} \\[10pt]\text{Given:} \\[4pt]L_1 = 20\,H \\[4pt]L_2 = 320\,H$$

Now, calculate the maximum possible mutual inductance:

​​​​$$\\[4pt]M = \sqrt{20 \times 320} \\[4pt]M = \sqrt{6400} \\[4pt]M = 80\,H$$

Therefore, the maximum possible mutual inductance between the coils is 80 H.

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In commercial buildings, lenses are used to distribute light evenly, which reduces brightness. 
Lenses can refract light, spreading it out over a wider area, which results in a more diffused and less intense light, effectively lowering the perceived brightness. 
Spot light: A spotlight is designed to concentrate light into a focused beam, the opposite of what lenses are used for in this scenario. 
Decoration: Decoration is not directly related to the function of lenses in lighting. 
Colour: Lenses primarily affect brightness and distribution of light, not its color. 

$$\text{Given:} \\[4pt]\bullet \; \text{Type of connection: Delta} \\[4pt]\bullet \; \text{Line voltage } V_L = 230\,\text{V} \\[4pt]\bullet \; \text{Phase current } I_p = 15\,\text{A} \\[4pt]\bullet \; \text{Power factor angle } \phi = 30^\circ \\[10pt]\text{For a delta-connected system, active power is given by:} \\[4pt]P = 3 \times V_p \times I_p \times \cos \phi \\[8pt]\text{(Since in delta, } V_p = V_L \text{)} \\[6pt]P = 3 \times 230 \times 15 \times \cos(30^\circ) \\[6pt]P = 3 \times 230 \times 15 \times 0.866 \\[6pt]P = 8.963\,\text{kW}$$​​

In a three-phase system, the phase displacement between the phases is 360°/3 = 120°, and the instantaneous power is constant, equal to three times the power per phase.

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