SSC JE Basic Electrical Questions with Detailed Solutions

$$\textbf{Concept:} \; \text{Energy consumed is the product of power and time:} \\[6pt]E = P \times t = I^2 \times R \times t \\[12pt]\text{Where:} \\\bullet \; E = \text{Energy in joules} \\\bullet \; I = \text{Current in amperes} \\\bullet \; R = \text{Resistance in ohms} \\\bullet \; t = \text{Time in seconds} \\[12pt]\textbf{Calculation:} \\[6pt]\text{Given:} \\\bullet \; R = 5 \, \Omega \\\bullet \; I = 20 \, A \\\bullet \; t = 10 \, \text{minutes} = 600 \, \text{seconds} \\[12pt]\text{Now,} \\E = (20)^2 \times 5 \times 600 = 400 \times 5 \times 600 = 1,200,000 \, J = 1200 \, kJ$$​​

The resistance of a wire is directly proportional to its length. According to the formula for resistance,

​$$R = \rho \frac{L}{A}$$​​

where RRR is the resistance, ρ is the resistivity of the material, L is the length, and A is the cross-sectional area of the wire.

As the length L increases, the resistance R increases as well, making them directly proportional.

$$\textbf{Concept:} \; \text{The power dissipated by a resistor is calculated using:} \\[6pt]P = I^2 \times R \\[12pt]\text{Where:} \\\bullet \; P = \text{Power in watts} \\\bullet \; I = \text{Current through the resistor} \\\bullet \; R = \text{Resistance in ohms} \\[12pt]\textbf{Calculation:} \\[6pt]\text{Given:} \\\bullet \; \text{Total current } = 40 \, A \\\bullet \; \text{Resistors in parallel: } 4 \Omega \text{ and } 6 \Omega \\\bullet \; \text{Total resistance in parallel:} \\[6pt]R_{eq} = \dfrac{4 \times 6}{4 + 6} = \dfrac{24}{10} \\[12pt]\text{Use current division rule to find current through 6}\Omega \text{ resistor:} \\[6pt]I_{6\Omega} = 40 \times \dfrac{4}{4+6} = 40 \times \dfrac{4}{10} = 16 \, A \\[12pt]P = (16)^2 \times 6 = 256 \times 6 = 1536 \, W$$​​

$$\text{The polar form is represented as } a \angle \theta, \text{ where:} \\[6pt]a = 10 \\\theta = 30^\circ \\[12pt]\text{To convert to rectangular form, the formulas are:} \\[6pt]x = a \cos(\theta) \\y = a \sin(\theta) \\[12pt]\text{Where:} \\x \text{ is the real part.} \\y \text{ is the imaginary part.} \\[12pt]\text{Using the values:} \\[6pt]x = 10 \cos(30^\circ) = 5\sqrt{3} \\y = 10 \sin(30^\circ) = 5 \\[12pt]\text{Thus, the rectangular form is:} \\[6pt]\text{Rectangular form } = 5\sqrt{3} + j5$$​​

$$\textbf{Concept:} \\[6pt]\text{The power factor is the ratio of active power to apparent power, given as:} \\[6pt]\cos \phi = \dfrac{P}{S} \\[12pt]\text{Where:} \\\bullet \; P = \text{Active Power} \\\bullet \; Q = \text{Reactive Power} \\\bullet \; S = \text{Apparent Power} \\[12pt]\textbf{Explanation:} \\[6pt]\text{In a purely resistive load (as given in the question), the current and voltage are in phase.} \\\text{This means:} \\[6pt]\bullet \; \text{The power factor } \cos \phi = 1 \\\bullet \; \text{Reactive Power } Q = 0, \text{ since resistors do not store energy} \\[12pt]\text{Therefore,} \\[6pt]\dfrac{Q}{P} = \dfrac{0}{P} = 0$$​​

$$\text{When } n \text{ identical resistors of resistance } R \text{ are connected in parallel, the total (equivalent)} \\\text{resistance is given by:} \\[8pt]\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R} + \cdots + \dfrac{1}{R} = \dfrac{n}{R} \;\;\Rightarrow\;\; R_{eq} = \dfrac{R}{n} \\[12pt]\text{So, the total resistance is } R \text{ divided by } n.$$​​

$$\text{To find the capacitance (C), we can use the formula:} \\[8pt]C = \dfrac{Q}{V} \\[12pt]\text{Where:} \\\bullet \; Q \text{ is the charge (1 coulomb),} \\\bullet \; V \text{ is the potential difference (1 volt).} \\[12pt]\text{Substituting the values:} \\[6pt]C = \dfrac{1C}{1V} = 1 \, \text{farad} \\[12pt]\text{Thus, the capacitance will be 1 farad.}$$​​

$$\textbf{Concept:} \\[6pt]\bullet \; \text{For capacitors in parallel, the total capacitance } C_{eq} \text{ is the sum of the individual capacitances:} \\[6pt]C_{eq} = C_1 + C_2 + \cdots + C_n \\[12pt]\bullet \; \text{For capacitors in series, the total capacitance } C_{eq} \text{ is given by:} \\[6pt]\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots + \dfrac{1}{C_n} \\[12pt]\textbf{Calculation:} \\[6pt]\bullet \; \text{The two 1F capacitors are in parallel:} \\[6pt]C_1 = 1F + 1F = 2F \\[12pt]\bullet \; \text{The resulting 2F capacitance is in series with the 6F capacitor:} \\[6pt]\dfrac{1}{C_2} = \dfrac{1}{2F} + \dfrac{1}{6F} = \dfrac{1}{1.5F} \;\;\Rightarrow\;\; C_2 = 1.5F \\[12pt]\bullet \; \text{The 1F and 2F capacitors are in parallel:} \\[6pt]C_3 = 1F + 2F = 3F \\[12pt]\bullet \; \text{The resulting 1.5F and 3F capacitors are in series:} \\[6pt]\dfrac{1}{C_{eq}} = \dfrac{1}{3F} + \dfrac{1}{1.5F} = \dfrac{1}{1F} \;\;\Rightarrow\;\; C_{eq} = 1F$$​​

$$\bullet \; \text{In an AC circuit, average power (P) is given by:} \\[6pt]P = V_{rms} \times I_{rms} \times \cos \theta \\[12pt]\text{where:} \\\bullet \; V_{rms} : \text{Root mean square voltage} \\\bullet \; I_{rms} : \text{Root mean square current} \\\bullet \; \theta : \text{Phase angle between voltage and current} \\\bullet \; \cos \theta : \text{Power factor}$$​​

​$$\bullet \;\; \text{Admittance } Y = G + jB \text{ where } B \text{ is susceptance.} \\[6pt]\bullet \;\; \text{For parallel elements in admittance form:} \\[6pt]B = j \left( \omega C - \dfrac{1}{\omega L} \right) \\[12pt]\text{Given:} \\[6pt]\bullet \;\; \omega = 16 \\ \bullet \;\; C = \dfrac{1}{8} \, F \\ \bullet \;\; L = \dfrac{1}{4} \, H \\[12pt]\text{Calculation:} \\[6pt]B = j \left( 16 \times \dfrac{1}{8} - \dfrac{1}{16 \times \tfrac{1}{4}} \right) \\[10pt]B = j \left( 2 - \dfrac{1}{4} \right) = j \cdot \dfrac{7}{4}$$​​

$$\bullet \; \text{Energy consumed (E) is calculated using the formula:} \\[8pt]E = P \times T \\[12pt]\text{where} \\[6pt]P = \text{Power in kilowatts (kW)}, \quad T = \text{Time in hours (h)} \\[8pt]1 \, \text{kW} = 1000 \, \text{watts (W)} \\[8pt]\text{For multiple devices, multiply the power of one device by the number of devices and the time of use.} \\[14pt]\textbf{Given:} \\[6pt]\bullet \; \text{Power of one bulb} = 100 \, W \\[4pt]\bullet \; \text{Number of bulbs} = 10 \\[4pt]\bullet \; \text{Operating time} = 20 \, \text{hours per day} \\[4pt]\bullet \; \text{Days in a month} = 30 \\[14pt]\textbf{Calculation:} \\[8pt]\text{Step 1: Total power of 10 bulbs} \\[6pt]10 \times 100 \, W = 1000 \, W = 1 \, kW \\[12pt]\text{Step 2: Energy consumed in one day} \\[6pt]E_{day} = 1 \, kW \times 20 \, \text{hours} = 20 \, kWh \\[12pt]\text{Step 3: Energy consumed in 30 days} \\[6pt]E_{month} = 20 \times 30 = 600 \, kWh$$​​

$$\textbf{Reactive Power in a Three-Phase System:} \\[10pt]Q = \sqrt{3} \cdot V_L \cdot I_L \cdot \sin \phi \\[14pt]\textbf{Where:} \\[6pt]\bullet \; V_L = \text{Line Voltage} \\[4pt]\bullet \; I_L = \text{Line Current} \\[4pt]\bullet \; \phi = \text{Power factor angle (between voltage and current)}$$​​

​$$\textbf{Given Data:} \\[6pt]\bullet \; \text{Line Voltage } V_{\text{line}} = 100 \; V \\[4pt]\bullet \; \text{Line Current } I_{\text{line}} = 40 \; A \\[4pt]\bullet \; \text{Phase angle } \phi = 36.87^\circ \\[10pt]\textbf{Formula Used:} \\[6pt]Q = \sqrt{3} \times V_{\text{line}} \times I_{\text{line}} \times \sin \phi \\[12pt]\textbf{Solution:} \\[6pt]\sin(36.87^\circ) = 0.6 \\[8pt]Q = \sqrt{3} \times 100 \times 40 \times 0.6 \\[8pt]Q = 1.732 \times 100 \times 40 \times 0.6 = 4156.8 \; \text{VAR} \\[8pt]Q = \frac{4156.8}{1000} = 4.156 \; \text{kVAR} \; \approx 4.155 \; \text{kVAR}$$​​

$$\text{We know that:} \\[6pt]\bullet \; 1 \, \text{ampere} = 1 \, \text{coulomb / second} \\[4pt]\bullet \; \text{Charge of one electron } = 1.6 \times 10^{-19} \, \text{coulombs} \\[10pt]\text{To find the number of electrons corresponding to 1 coulomb of charge:} \\[8pt]\text{Number of electrons} = \frac{1C}{1.6 \times 10^{-19} C / \text{electron}} = 6.25 \times 10^{18} \, \text{electrons} \\[12pt]\text{So, 1 ampere of current implies } 6.25 \times 10^{18} \text{ electrons flow per second.}$$​​

A pure resistive device means that it primarily converts electrical energy into heat with minimal energy loss to other forms like light or mechanical energy. A heater is designed specifically to produce heat, making it the correct answer.