SSC JE Basic Electrical Questions with Detailed Solutions

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

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

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.

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

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

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

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

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

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

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

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

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. 

$$\text{Step 1: Find the equivalent resistance of } R_2 \text{ and } R_3 \text{ in parallel:} \\[8pt]\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \;\Rightarrow\; R_{23} = \frac{6}{5} = 1.2 \, \Omega \\[12pt]\text{Step 2: Add } R_1 \text{ in series with } R_2 \parallel R_3 : \\[8pt]R_{\text{total}} = R_1 + R_{23} = 1 + 1.2 = 2.2 \, \Omega$$​​

Concept of Temperature Coefficient of Resistance:

The temperature coefficient of resistance (α) tells us how much the resistance of a material changes per degree change in temperature. The relation is given by:​​

​$$\\[10pt]R_t = R_0 \left( 1 + \alpha (T - T_0) \right) \\[14pt]\textbf{Where:} \\[6pt]\bullet \; R_0 = \text{Resistance at reference temperature } T_0 \\[4pt]\bullet \; R_t = \text{Resistance at temperature } T \\[4pt]\bullet \; \alpha = \text{Temperature coefficient of resistance} \\[4pt]\bullet \; T_0 = \text{Initial temperature} \\[4pt]\bullet \; T = \text{Final temperature} \\[14pt]\textbf{Given Data:} \\[6pt]\bullet \; R_0 = 20 \, \Omega \; \text{at } T_0 = 20^\circ C \\[4pt]\bullet \; R_t = 40 \, \Omega \; \text{at } T = 60^\circ C \\[14pt]\textbf{Calculation:} \\[6pt]\text{Using the formula:} \\[6pt]R_t = R_0 \left( 1 + \alpha (T - T_0) \right) \\[12pt]40 = 20 \left( 1 + \alpha (60 - 20) \right) \\[8pt]40 = 20 (1 + 40\alpha) \\[8pt]2 = 1 + 40\alpha \\[8pt]1 = 40\alpha \\[8pt]\alpha = \frac{1}{40} \; \text{per}^\circ C$$​​

$$\bullet \; \text{Voltage across the capacitor:} \\[6pt]V(t) = 10t^3 - 5t + 10 \\[12pt]\bullet \; \text{Capacitance:} \quad C = 10 \, F \\[12pt]\text{We are to find the current through the capacitor at } t = 2 \, s. \\[12pt]\textbf{Formula Used:} \\[6pt]\text{The current through a capacitor is:} \\[6pt]i(t) = C \cdot \frac{dV(t)}{dt} \\[12pt]\textbf{Given:} \\[6pt]V(t) = 10t^3 - 5t + 10 \\[12pt]\text{Differentiate } V(t) \text{ with respect to } t: \\[6pt]\frac{dV(t)}{dt} = 30t^2 - 5 \\[12pt]\text{Now plug into the current formula:} \\[6pt]i(t) = 10 \cdot (30t^2 - 5) = 300t^2 - 50 \\[12pt]\text{Now substitute } t = 2: \\[6pt]i(2) = 300 \cdot (2)^2 - 50 = 300 \cdot 4 - 50 = 1200 - 50 = 1150 \, A$$​​