- Basic Concepts of Electrostatics
- How does electricity work?
- Faraday’s Laws of Electromagnetic Induction
- Ideal & Practical Voltage Sources
- Internal Resistance of Electric Source
- Ohm’s Law and Concept of AC & DC
- What is Peak, R.M.S. & Average Value of ac voltage and current
- Kirchhoff’s Current Law (KCL)
- Kirchhoff’s Voltage Law (KVL)
- Concept of Electric Power
- Concept of AC voltage – Learn in simple language
Definition: The rate of doing electric work is called electric power. The unit of electric power is Joule/second or Watt (W).
The electric power can also be measured in kilo-watt-hour (kWh) or Joules (J). It is given as follows:
1kWh = 1000Watt – hour
= 1000 x 3600 Watt – seconds or Joules
= 1000 x 3600 Watt – seconds or Joules
If you solve numerical problems on DC electric power given below your concept will be more clear. Remember the simple definition of DC power is as follows:
Definition: The DC power dissipated in a device is the product of its operating voltage and the amount of current flowing through it.
P = I.V = I2.R = V2/R
Exercise
- Calculate the power of an electric bulb, if its current is 1.5A at 220V. (Ans: 330W)
- In a small house, two tubes of 40W each, one TV of 100W and three fans of 60W each are used. Calculate the total current required from AC supply from MSEB of 220V, when all these loads are switched on. (Ans: 1.64A)
- The electric power dissipated in a DC operated bulb is 100W at 12V. Then calculate the amount of current flowing through the bulb. (Ans: 8.33A)
- The mobile charger produces a constant voltage of 3.7V during charging. The mobile battery has a capacity of 2000mA. If the mobile charger delivers 500mA charging current, in how much time the mobile battery will be fully charged? (Ans: 4 hours)
- Calculate the power delivered by mobile phone battery, if it can produce a maximum load current of 1200mA at 3.7V. (Ans: 4.44W)
- In the following circuit, the unknown resistor (R) dissipates 12W of power from 16V battery. Find the current (Ix) and value of (R). (Ans: Ix = 1A & R = 12W or Ix = 3A & R = 1.33W)
- Using KVL, calculate the voltage drop across each resistor in following circuit, if R1=R4=2Ω, R2=5Ω, R3=R5=3Ω. (Ans: V1=V4=1.33V, V2=3.33V, V3=V5=3V)
