1 Crore+ students have signed up on EduRev. Have you? 
Power is the product of voltage and current. Writing I in terms of V, we get P=V^{2}/R and writing V in terms of I, we get P=I^{2}r.
Power is proportional to both voltage and current. Hence both the options are right.
A 250V bulb passes a current of 0.3A. Calculate the power in the lamp.
Here, V = 250v and I = 0.3A. P=VI. Which implies that, P=250*0.3=75W.
Power is the energy per unit time. That is, P=E/t. If the unit of power is kW and the unit of time is hour, then the unit of energy=unit of power*unit of time=kWh.
Here V = 200v and Resistance( R) = 20ohm. P=V^{2}/R= 200^{2}/20=2000W=2kW
A current of 5A flows in a resistor of 2 ohm. Calculate the power in the resistor and the energy dissipated in 300 seconds.
P = I^{2}R = 5^{2}*2 = 50W.
E = Pt = 50*300 = 15000J = 15kJ.
This is a series connected circuit hence the current across each resistance is the same. To find current: I=V/R=200/40=5A.
To find power: P=I^{2}R=(5)^{2}*20 = 500W. Since both the resistors have a resistance of 20 ohm, the power across both is the same.
This parallel connected circuit, hence the voltage across each of the resistors are the same. P=V2/R= 1002/10=1000W=1kW. Since both the resistors receive the same amount of voltage, the Power in both are the same.
Calculate the work done in a resistor of 20 ohm carrying 5A of current in 3 hours.
To find power: P=I^{2}R=5^{2}*20=500W=0.5kW.
To find Work done: W=Pt=0.5*3=1.5kWh.
Power = energy/time =J/s(joules per second). Hence the Si unit of Power is J/s.
Watt is a unit of power and hour is a unit of time. Energy is the product of power and time, hence the unit for power is kWh.
A bulb has a power of 200W. What is the energy dissipated by it in 5 minutes?
Here, Power = 200w and time = 5min. E=Pt => E= 200*5= 1000Wmin=60000Ws= 60000J= 60kJ.
Out of the following, which one is not a source of electrical energy?
A potentiometer is an instrument used for measuring voltage hence it is not a source for electrical energy.
Here V = 100 and R = 10. Power in the circuit= V^{2}/R= 100^{2}/10= 1000W.
Energy= Pt= 1000*50= 50000J= 50kJ.
Expression for power= VI, substituting I from ohm’s law we can write, P=V^{2}/R. Energy is the product of power and time, hence E=Pt= V^{2}t/R.
Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V^{2}/R. P=200^{2}/10= 4000W. E=Pt= 4000*10=40000Ws= 40000J= 40kJ.
A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy.
A current of 2A flows in a wire offering a resistance of 10ohm. Calculate the energy dissipated by the wire in 0.5 hours.
Here I (current) = 2A and Resistance(R) = 10ohm. Power= I^{2}R= 2^{2}*10=40. Energy=Pt= 40*0.5*60*60= 72000J=72kJ.
The current in the circuit is equal to the current in the 5 ohm resistor since it a series connected circuit, hence I=220/(5+10)=14.67A. P=I^{2}R= 14.67^{2}*5=1075.8W. E=Pt= 1075.8*20= 21516J=21.5kJ.
Practically, if 10kJ of energy is supplied to a device, how much energy will the device give back?
Practically, if 10kJ of energy is supplied to a system, it returns less than the supplied energy because, some of the energy is lost as heat energy, sound energy etc.
57 docs59 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
57 docs59 tests









