Test: Heating Effect of Electric Current - Class 10 MCQ

If the current passing through an electric heater has been halved then the heat produced will also be reduced to one fourth because H=I² RT.

To calculate the number of days an 18-watt bulb consumes 1 unit of electric energy when used for one hour daily across a 240 V power line:

The energy consumed by the bulb in one hour is:

• Energy (in watt-hours) = Power (in watts) × Time (in hours)
• For the bulb:
  • Power = 18 watts
  • Time = 1 hour
  • Energy consumed = 18 W × 1 h = 18 Wh
• 1 unit of electric energy is equivalent to 1 kWh, which is 1000 Wh.
• To find the number of days (n) to consume 1 unit:
  • Energy consumed in n days = 18 Wh × n
  • Set this equal to 1000 Wh:
  • 18n = 1000
  • n = 1000 / 18 ≈ 55.56 days

Thus, the bulb will consume 1 unit of electric energy in approximately 55.5 days

We know that, R = V² /P

Hence resistance of 1st bulb = (250)² / 200 = 625/2

resistance of 2nd bulb = (250)² / 100 = 625

Total resistance of the circuit when the bulbs are connceted in series = 625/2 + 625 = 1875/2 ohm

Therefore, total power consumed in the circuit , P = (250)² / (1875/2) = 67 watt

Electric power in terms of current = IR² and in terms of voltage = V²/R and in terms of voltage and current both = VI

From ohm’s law, I = V/R = 220/440 = 0.5 A

H = I² × R × t = (0.5)² × 440 × 10 = 1100 J

To calculate the cost of operating a refrigerator for 30 days, follow these steps:

• The refrigerator uses 500 W or 0.5 kW.
• It runs for 10 hours daily.
• Total energy used in 30 days: 0.5 kW × 10 hours × 30 days = 150 kWh.
• Cost per kWh is Rs. 4.00.
• Total cost: 150 kWh × Rs. 4.00 = Rs. 600.

The Fuse wire connection are in between the appliance and main line. It is are very thin wire and when very high current passes through it suddenly, the fuse wire gets heated and melts due to the produced heat. This saves the household appliances

To find the potential difference across the resistor, use these steps:

• Power is given as 125 J/s, which is equivalent to 125 watts.
• The resistance of the resistor is 5 ohms.
• Use the formula for power: P = V² / R.
• Rearrange to find the voltage: V = √(P × R).
• Substitute the values: V = √(125 × 5).
• Calculate: V = √625 = 25 V.

Therefore, the potential difference is 25 volts.

The power of the bulb can be calculated using the formula:

• Given Voltage (V) = 270 Volts
• Given Current (I) = 0.5 Amps
• Power (P) = Voltage × Current
• Therefore, P = 270 × 0.5 = 135 watts

P = I² R = (10)² × 5 = 500 watt = 0.5 kW

t = 30 minute = 0.5 hr

E = P × t = 0.5 kW × 0.5 hr = 0.25 kWh

Power = voltage x current 
=  Voltage x ( voltage / resistance) 
Hence, if P is power, V is Voltage and R is resistance, then P = V ²/ R 
Hence for first bulb, R1 = V²/75 = R
For second bulb, R2 = V² / 150
 = > ½ (V ²/75) = R/2 
=> R/2

The resistance of a wire is determined by its length, material, and cross-sectional area.

• Resistance is inversely proportional to the cross-sectional area. This means as the area increases, the resistance decreases.
• If the diameter of a wire is doubled, the cross-sectional area increases by a factor of four, since area is proportional to the square of the diameter.
• Therefore, the new resistance becomes one-fourth of the original resistance.

Thus, the resistance of the second wire is 0.25 R.

The commercial unit of electrical energy is the kilowatt-hour (kWh).The commercial unit of electrical energy is kilowatt hour.