- 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
Every electric source like battery or generator contains some internal resistance. This resistance is invisible, but it is always present within the source. If the internal resistance of the battery or generator is very low, its power delivering capacity is very high.
In following diagram a battery is shown with its internal resistance of r=5Ω. Across the terminals of battery A-B the load resistor of RL=10Ω is connected. Now we shall see how internal resistance (r) of the battery affects its working.
The e.m.f. of the battery is: E = 12V
So the total current (I) delivered by the battery is: I = E/R = 12/15 = 0.8A
This current flows through the internal resistance (r) first and then through the load resistor RL. So the voltage drop across them will be:
VRL = I.RL = 0.8 x 10Ω = 8V
Now the output voltage (PD) of the battery will be only 8V, because 4V will be dropped inside the battery within its internal resistance (r).
So we can conclude that greater is the internal resistance of the battery, smaller is its output voltage.
Exercise
- Calculate the PD of the battery connected in a circuit, if its internal resistance is 20Ω and e.m.f. is 120V. Consider that a load resistance of 100Ω is connected across the battery terminals.
(Ans: PD = 100V, I = 1A) - In a circuit, if r = 1.5W and current in the circuit is 0.1A with load voltage = 60V, calculate the e.m.f. of the battery and value of load connected in the circuit. (Ans: E = 60.15V, RL = 600Ω)
- Calculate the total current delivered by the battery with internal resistance of 1Ω and load resistance of 11Ω, if its e.m.f. is 24V. Also calculate the PD of the battery. (Ans: I = 2A, PD = 22V)