Quiz: Electrical Engineering
Exam: UPPSC AE
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. A 6 A source has a shunt internal resistance of 85 Ω. The maximum power that can be delivered to a load is
(a) 475 W
(b) 765 W
(c) 585 W
(d) None of above
Q2. Fuse wire is a wire of
(a) high R and low MP
(b) high R and high MP
(c) low R and low MP
(d) low R and high MP
Q3. Two electric bulbs whose resistances are in the ratio 1 : 2 are connected in parallel to a constant voltage source. The powers dissipated by them are in the ratio
(a) 1 : 2
(b) 1 : 1
(c) 2 : 1
(d) 4 : 1
Q4. The current in a 484 W, 220 V heater operated at 200 V is
(a) 0.75 A
(b) 3/8 A
(c) 2 A
(d) 15/4 A
Q5. Two lamps of 100 W and 240 W rated for 220 V are placed in series and a 400 V applied across them. Then
(a) only 100 W lamp will fuse
(b) only 200 W lamp will fuse
(c) both the lamps will fuse
(d) no lamp will fuse
Q6. A coil of wire of resistance 50 Ω is embedded in a block of ice. What is the amount of ice melted in 1 s if a potential difference of 210 V is applied across the coil?
(a) 2.625 g
(b) 3.25 kg
(c) 4.12 kg
(d) None of the above
Q7. What is the maximum safe current flow in a 36 Ω, 4 W resistor?
(a) 0.47 A
(b) 0.39 A
(c) 0.57 A
(d) None of the above
Q8. Which of the following parameter will remain constant during charging and discharging of a capacitor?
c) Time constant
Q9. What is the value of initial current while charging a capacitor?
d) Cannot be determined
Q10. A capacitor is charged to a voltage of 440V and has a resistance of 40 ohm. Calculate the initial value of charging current.
Sol. Here, I_N = 6 A and R_N = 85 Ω. For maximum power in the load, R_L=R_N = 100 Ω.
Therefore, current through load = I_N/2 = 6/2 = 3 A
∴ P_max = (3)² × R_L = (3)² × 85 = 765 W
Sol. Fuse wire should have High R and Low MP
Sol. 2 : 1 Due to inverse relation.
Sol. Resistance, R = V^2/P=(220)^2/484 = 100 Ω
Sol. Resistance of 100 W lamp, R₁ = V^2/P_1 =(220)^2/100 = 484 Ω
Resistance of 240 W lamp, R₂ = V^2/P_2 =(220)^2/240 = 201.6 Ω
Total circuit resistance, R_T = R₁ + R₂ = 484 + 201.6 = 685.66 Ω
Circuit current, I = 400/685.66 = 0.583 A
Voltage across 100 W lamp, V₁ = I R₁ = 0.583 × 484 = 282.35
Voltage across 240 W lamp, V₂ = 400 – 282.35 = 117.6 V
Note that voltage across 100 W lamp is more than its rated voltage while voltage across 240 W lamp is less than its rated voltage. Therefore, only 100 W lamp will fuse.
Sol. Heat produced per second,
H=1/4.2×V^2/R cal=1/4.2×(210)^2/50=210 cal
Now, H = mL where L is latent heat of fusion of ice = 80 cal/g
Or m = H/L=210/80 = 2.625 g
Sol. P = I²R or I = √(P/R)=√(4/36) = 0.574A
Explanation: The time constant in a circuit consisting of a capacitor is the product of the resistance and the capacitance (which are fixed for a particular capacitor). Smaller the time constant, faster is the charging and discharging rate and vice versa.
Explanation: The initial current of a capacitor is very high because the voltage source will transport charges from one plate of the capacitor to the other plate.
Explanation: Initially, there’s 0V voltage in a capacitor. As the capacitor charges, the voltage increases. Since a voltage is proportional to current by ohm’s law, initial current is also equal to zero.