Advertisements
Advertisements
प्रश्न
A bulb is connected to a battery of p.d. 4 V and internal resistance 2.5 Ω . A steady current of 0.5 A flows through the circuit. Calculate:
The energy dissipated in the bulb in 10 minutes.
उत्तर
Given ,
Voltage , V = 4 V
Resistance of the battery , RB = 2.5 Ω
Current , I = 0.5 A
Energy dissipated in the bulb in 10 min , E = I2Rt
E = (0.5)2 × 5.5 × 600 = 825 J
APPEARS IN
संबंधित प्रश्न
What actually travels through the wires when you switch on a light?
Why should the resistance of:
an ammeter be very small?
Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.
Name two devices which work on the heating effect of electric current.
Two resistors of 4 Ω and 6 Ω are connected in parallel. The combination is
connected across a 6 volt battery of negligible resistance. Calculate:
( i) The power supplied by the battery,
(ii) The power dissipated in each resistor.
A 3 pin mains plug is fitted to the lead for a 1 kW electric kettle to be used on a 250 V A.C. supply. Which of the following statements is NOT correct?
(a) A 13 A fuse is the most appropriate value to use.
(b) The brown wire should be connected to the live side of the mains.
Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply of the energy supplied in kWh to the three heaters in 5 hours.
Explain, why the potential difference across the terminals of a cell is more when the cell is not in use than it is when the cell is being used.
An electrical appliance is rated at 1000 KVA, 220V. If the appliance is operated for 2 hours, calculate the energy consumed by the appliance in: (i) kWh (ii) joule.