Railway Science (Physics-Electric Current and its Effects) (Part-VIII)

Total Questions: 50

21. What is the total resistance of three 20 Ω coils connected in series? [RRB Group D 27/09/2018 (Evening)]

Correct Answer: (d) 60 Ω
Solution:

Given : Resistance of three 20 Ω coils
R₁ , R₂ , R₃ …… in Series
= R₁ + R₂ + R₃ + …….. So, 20 Ω + 20 Ω + 20 Ω = 60 Ω.

22. An electric bulb of 100 W is used 8 hours per day. The energy consumed by the bulb in one day is _____ units. [RRB Group D 28/09/2018 (Morning)]

Correct Answer: (d) 0.8
Solution:

Given, Power = 100 W, Time, t = 8 hours
Energy consumed by bulb in one day = 8 × 100W = 800 Wh
We know, 1 kWh=1000 Wh = 1 unit
Hence 800 Wh = 0.8 unit for 1 day.

23. An electric bulb is connected to a 220 V generator. The current through it is 0.50 A. What is the power of the bulb ? [RRB Group D 28/09/2018 (Evening)]

Correct Answer: (d) 110 W
Solution:

Here, the potential of the generator = 220 V, Current flowing across the electric bulb, I = 0.50 A.
∴ Power of the bulb, P = V × I = 220 × 0.50 = 110

24. When an electric current of 4A flows from a heater source, the kinetic difference between its terminals is 60V. How much current does the heater draw if the dynamic difference is increased to 157.5 V? [RRB Group D 01/10/2018 (Morning)]

Correct Answer: (a) 10.5 A
Solution:

According to Ohm’s Law, resistance (𝑅) = 𝑉/I = 60/4 = 15Ω.
Now voltage (V) is increased to 157.5 V, then
Current (I) = 𝑉/R  = 157.5/15 = 10.5 A.

25. An electric current of 2 A flows through a conductor whose two ends are at a kinetic energy of 4 V. The resistance of a conductor will be : [RRB Group D 01/10/2018 (Morning)]

Correct Answer: (c) 2 ohm
Solution:

Given that, voltage (V)= 4V and Current (I) = 2A.
R =  𝑉/I = 4/2 = 2Ω.

26. In an electric heater, when an electric current of 4 A flows from a source, its potential difference is 60 V. If the potential difference increases to 165 V, then what will be the electric current? [RRB Group D 01/10/2018 (Afternoon)]

Correct Answer: (d) 11 A
Solution:

Given that, Potential difference = 60V,
Current drawn = 4A, Resistance = ?
According to Ohm law, V = IR
R =  𝑉/I, R = 60/4 = 15 ohms.
Now, Potential difference = 165V, Resistance = 15 ohms, Current = ?
V = RI, I = 𝑉/R = 165/15 = 11 A.

27. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 172.5 V? [RRB Group D 01/10/2018 (Evening)]

Correct Answer: (d) 11.5 A
Solution:

Given, V = 60 V, I = 4 A
Ohm’s Law : V = I × R
So, R = V/I = 60/4 = 15  Ω
When potential difference is increased to 172.5V
I = V/R = 172.5 = 11.5 A.

28. An electric iron of 750 W is used for 8 hours per day. Then the energy consumed in one day by the iron is: [RRB Group D 03/10/2018 (Morning)]

Correct Answer: (a) 6 units
Solution:

Given, Power (P) = 750 W = 0.75 kW, Time (t) = 8 hours.
Energy Consumed = (P × t ) = (0.75 × 8) = 6 kWh = 6 units.

29. The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 127.5 V ? [RRB Group D 03/10/2018 (Morning)]

Correct Answer: (a) 8.5 A
Solution:

Given, Potential Difference (V) = 60 V,
Current (I) = 4 A, V = I × R (By Ohm’s law).
Resistance (R) = V/I  = 60/4 = 15 ohm.
Now, New Potential Difference = 127.5 V.
Current (I) = V/R = 127.5/15 = 8.5 A.

30. A 5 Ω resistance wire is doubled on it. Calculate the new resistance of the wire. [RRB Group D 03/10/2018 (Afternoon)]

Correct Answer: (a) 1.25 Ω
Solution: