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

Total Questions: 50

21. The current (I) flowing through an electric circuit of resistance 10 ohm is 5 amp. What is the heat generated per second? [RRC Group D 26/09/2022 (Afternoon)]

Correct Answer: (a) 250 joule
Solution:

According to Joules law of Heating, H = I²Rt. Where I is current, R is resistance and t is time. Given R = 10 Ω, I = 5 A. Then heat generated per second is H = 5² × 10 × 1 = 250 joule.

22. A battery of 12V supplies a current of 3 A to a lamp connected to it. The energy supplied to it in 15 minutes will be: [RRC Group D 26/09/2022 (Evening)]

Correct Answer: (c) 32400 J
Solution:

The energy supplied to the lamp can be calculated using the formula:
Energy = Power ×  Time.
Now, Power = Voltage × Current = 12V × 3A = 36W.
The time for which the energy is supplied is 15 minutes, which is equivalent to 900 seconds. Substituting these values in the formula for energy, we get: Energy = Power × Time = 36W × 900s = 32,400 J

23. If the current flowing through a resistance of 5 ohm is 6 amp, then the current flowing through another resistance of 10 ohm connected in series to it is_________ [RRC Group D 27/09/2022 (Morning)]

Correct Answer: (d) 6 amp
Solution:

In parallel connection, Voltage remains constant throughout the circuit while for series connection, Current remains constant throughout the circuit.

24. The potential difference across a resistor of 2 ohm when 200 joule of heat is produced each second is given by_______. [RRC Group D 27/09/2022 (Evening)]

Correct Answer: (c) 20 volt
Solution:

Given Heat (H) = 200 J, time (t) = 1 second and Resistance = 2 Ω.

25. In an electrical circuit, five identical bulbs are connected in parallel to each other to a battery of 20 V (negligible internal resistance). When all the five bulbs glow, a current of 5 A is recorded. Then the power dissipated in the circuit and the resistance of each bulb are _______ and _______ respectively. [RRC Group D 27/09/2022 (Evening)]

Correct Answer: (b) 31.25 W, 1.25 Ω
Solution:

The resistance of one bulb is R.
So total resistance of 5 bulb = R/5
In the given electrical circuit
Given V = 20 v, Total current I = 5 A
According to ohm law V = IR
⇒ 20 = 5 R , R = 4 Ω
⇒ so resistance of each bulb Rₒₙₑ =
Total bulb/R = 5/4 = 1.25 Ω
Power in Circuit P = I²R ⇒ (5)² (1.25) = 31.25 W

26. The power of an electric equipment through which a charge of 15 coulomb is flowing per 5 seconds when it is connected to a 20 V source is ______. [RRC Group D 28/09/2022 (Afternoon)]

Correct Answer: (c) 60 watt
Solution:

Q= I × T
Where Q = charge per second, I = current,
T = time Q = 15 coulomb, T= 5 sec.
15 = I × 5
I = 3 A
Now, P = VI
Where P = power, V = voltage, I = current
P = 20 × 3 = 60 Watts.

27. Heating device of power 1100 W is designed to operate at 220 V line voltage. If on a particular day the line voltage drops to 110 V, then the current through it and its output power will be _______ and _______. [RRC Group D 29/09/2022 (Afternoon)]

Correct Answer: (d) 2.5 A, 275 W
Solution:

28. Find the resistance of the conductor using the given graph. [RRC Group D 29/09/2022 (Afternoon)]

Correct Answer: (d) 60 ohm
Solution:

From Ohm's law, the resistance of a conductor is given as V = IR
Hence, it is the slope of the V-I curve. So resistance at any point on the graph will be ⟹ R = 3 - 0 / 0.05-0.00 ⟹ 60 Ω

29. The net resistance of two resistors R₁ and R₂ connected in series is 8 ohm and their net resistance in parallel is 2 ohm. What are the values of individual resistances R₁ and R₂ , respectively? [RRC Group D 29/09/2022 (Evening)]

Correct Answer: (b) 4 ohm; 4 ohm
Solution:

30. The length of a conducting wire is 1 km and its radius of cross section is 7 mm. A resistor of resistance 10 Ω is made out of this wire. The resistivity of this conductor will be________ [RRC Group D 30/09/2022 (Afternoon)]

Correct Answer: (a) 1.54 × 10⁻⁶Ω-m
Solution: