SSC JE Electrical 2019 with solution SET-1 Find the net capacitance of the combination in which ten capacitors of 10 μF are connected in parallel 1 μF 100 μF 50 μF 0.1 μF 1 μF 100 μF 50 μF 0.1 μF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP When the capacitance is connected in series the total capacitance is given byCeq = C1 + C2 + C3………CNCeq = 10nCeq =10 × 10 = 100 μF
SSC JE Electrical 2019 with solution SET-1 Find the mutual inductance between two ideally coupled coils of 2H and 8H 4 H 8 H 16 H 2 H 4 H 8 H 16 H 2 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Mutual inductance between the two coils is K = √L1.L2K = √2 × 8k = √16k = 4 H
SSC JE Electrical 2019 with solution SET-1 The ratio of average energy demand to maximum demand during a specific period in Power factor Utilization factor Load factor Form factor Power factor Utilization factor Load factor Form factor ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The ratio of average energy demand to maximum demand during a specific period is called the load factor.Load Factor = Average Load/Maximum Demand
SSC JE Electrical 2019 with solution SET-1 In the circuit shown, if R = 0, then the phase angle between v(t) and i(t) is 30° 0° 90° 60° 30° 0° 90° 60° ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In the given circuit if R = 0 then the circuit becomes purely inductive. So the phase angle between v(t) and i(t) is 90°.
SSC JE Electrical 2019 with solution SET-1 When the only current source is active in the circuit, find the current through the 10Ω resistor 1.66 A 1.33 A 0.66 A 0 A 1.66 A 1.33 A 0.66 A 0 A ANSWER DOWNLOAD EXAMIANS APP
SSC JE Electrical 2019 with solution SET-1 The voltmeter is shown in the circuit reads: 1.2 V 24 V 12 V 2.4 V 1.2 V 24 V 12 V 2.4 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Total resistance in the given circuitR = (250 + 250)MΩ = 500 MΩCurrent I = V/R = 24/(500 × 103)Now the Voltage in the voltmeter= \dfrac{{24}}{{500 \times {{10}^3}}} \times 250 \times {10^3}V = 12 V
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