Advertisements
Advertisements
प्रश्न
Two resistors A and B of 4 Ω and 6 Ω respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate the power supplied by the battery.
Advertisements
उत्तर
Given,
Resistance, RA = 4 Ω
Resistance, RB = 6 Ω
Voltage, V = 6 V
Ω As resistances are connected in parallel
Equivalent Resistance = `1/"R" = 1/"R"_"A" + 1/"R"_"B"`
`1/"R" = 1/4 + 1/6`
= `10/24`
= `5/12`
R = 2.4 Ω
As power, P = `"V"^2/"R"`
= `(6)^2/2.4`
= 15 W
APPEARS IN
संबंधित प्रश्न
What determines the rate at which energy is delivered by a current?
Which particles are responsible for current in conductors?
Name two gases which are filled in filament type electric light bulbs.
An electric bulb is rated 250 W, 230 V. Calculate:
- the energy consumed in one hour
- the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains.
A battery of e.m.f 15 V and internal resistance 2 Ω is connected to two resistors of resistance 4 ohm and 6 ohm joined in series. Find the electrical energy spent per minute in 6 ohm resistor.
A resistor of resistance R is connected to an ideal battery. If the value of R is decreased, the power dissipated in the resistor will ______________ .
An electric bulb marked 220 V, 100 W will get fused if it is made to consume 150 W or more. What voltage fluctuation will the bulb withstand?
A bulb is connected to a battery of e.m.f. 6V and internal resistance 2Ω A steady current of 0.5A flows through the bulb. Calculate the total energy supplied by the battery in 10 minutes.
A consumer uses 4 lamps of 60 watt, 2 lamps of 40 watt, and 2 lamps of 100 watt. All these are used for 6 hour daily. Find the total bill for 30 days when the rate of energy is 75 paise per unit and the meter rent is Rs. 1.
A conductor of 10 Ω is connected across a 6 V ideal source. The power supplied by the source to the conductor is ______.
