Advertisements
Advertisements
Question
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.
Advertisements
Solution
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
RELATED QUESTIONS
Fill in the blank:
To increase the effective resistance in a circuit the resistors are connected in _____________.
The p.d. across a 3 Ω resistor is 6 V. The current flowing in the resistor will be:
(a) `1/2A`
(b) 1 A
(c) 2 A
(d) 6 A
Draw a circuit diagram to show how two 4 V electric lamps can be lit brightly from two 2 V cells.
Calculate the equivalent resistance between A and B in the adjacent diagram.
Define the following:
Super conductors
What is the purpose of fuse in an electrical circuit?
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.
Three resistances, of which two are equal when connected in series have an effective resistance of 30 ohms. When the three resistances are connected in parallel, the effective resistances is 3 ohms. Find the values of the individual resistances.
