 Hello everyone, myself Sanjay Uttgay, assistant professor, department of electronics engineering, Wailchand Institute of Technology, SolarPort. Today, we are going to discuss numericals on DC circuit part 2, learning outcome at the end of this session, students will be able to analyze and simplify complicated DC circuits, guidelines, review of electrical laws, numericals on DC circuits, assignment references, introduction. Ohm's law states that for a linear circuit, the current flowing through it is proportional to the potential difference across it. So, the greater the potential difference across any two points, the bigger will be the current flowing through it. For example, if the voltage at one side of a 10 ohm resistor measures 8 volt and at the other side of the resistor, it measures 5 volt, then the potential difference across the resistor will be 3, that is 8 minus 5 volts causing a current of 3 divided by 10 ohm, that is 0.3 ampere, potential difference. Here point A and point B, here at point A it is a plus 8 volt, at point B it is 5 volt. At point A it is a plus 8 volt, at point B it is 5 volt. So, the resistor on the other side of the resistor, one voltage is 8 volt and another is 5 volt. So, V AB will be the difference that is 8 minus 5 3 volts. Moving to the next resistance, again point A and point B, point A is at 6 volt and point B at which is at minus 2 volts. So, the net resistance which is the difference 6 minus of minus 2, so it will be 8 volts. Coming to the third resistance, here point A is at plus 3 volts, point B which is at minus 3 volts. So, the potential difference across this resistance will be 3 volt minus of minus 3 comes out to be V AB equals to 6 volts. Example number one, calculate the current flowing through a 100 ohm resistor that has one of its third terminal connected to 5050 volts and other terminal connected to 30 volts. Voltage at terminal A is 50 volt, at B it is 30 volt. So, voltage across the net voltage across the resistance will be 50 minus 30 that is 20 volts. The voltage across the resistor being 20 volt then the current flowing through the resistor is given by I is equal to V upon R that is I is equal to V AB divided by resistance that is 100 ohm comes out to be 20 volt divided by 100 ohm comes out to be 0.2 ampere that is 200 milli ampere. Example number two, in a given circuit 4 resistors of value R1 10, R2 20, R3 30 ohm, R4 40 ohm are connected across 100 volt DC supply. Calculate the voltage drop at points P 1, P 2, P 3 and P 4 and also the individual voltage drop across each resistors within the series chain. Voltage divided circuit, so what is voltage divided circuit? It is basically a series combination of number of resistances here where having R1, R2, R3 and R4 all in series connected to a single power source supply voltage V s whereas, points P 1, P 2, P 3 and P 4. P 1 will be the voltage across R1, P 2 will be the voltage across R1 plus R2 whereas, P 3 will be the summation of drop across R1 plus R2 plus R3 that is P 1 plus R2 plus V R1 plus V R2 plus V R3 and finally, P 4 is the summation of voltage drop across all resistances that is R1, R2, R3 and R4 that is V R1 plus V R2 plus V R3 plus V R4. The voltages at the various points are calculated as V P 1 is equal to R 1 upon R total into V s. So, what is R 1? It is 10. What is supply voltage? 100. What is total resistance? It is 10 plus 20 plus 30 plus 40 comes out to be 10 volts. Now, V P 2 is equals to what? P 2 point covers two resistances R 1 and R 2 in series. So, it will be V P 2 will be equals to R 1 plus R 2 into total supply voltage divided by total resistances. So, V P 2 comes out to be 10 plus 20 multiplied by that is 30 multiplied by 100 upon total resistances comes out to be 100. So, V P 2 is equals to 30 volts. V P 3 is the summation of voltage across R 1 and R 2 and R 3. So, R 1, R 2, R 3 they are in series. So, V P 3 will be equals to R 1 plus R 2 plus R 3. It is the series combination into V s upon R t which is equals to 10 plus 30 plus 30 is equal to 60 into 100 upon 100. 100 100 will get cancelled. So, it is 60 volts. Now, let us find out the individual voltage drops across each resistor. I is equals to supply voltage upon total supply voltage V s upon total resistances. Supply voltage is given equals to 100 volts. R total is equals to summation of all the resistances as all resistances R 1, R 2, R 3, R 4 all are in series. So, that total resistance will be 10 plus 20 plus 30 plus 40 which is equals to 100 ohm. So, total current comes out to be 100 volt upon 100 ohm comes out to be 1 ampere. This is the total current. Now, let us find out the individual voltage drop across R 1, R 2, R 3 and R 4. So, V R 1 will be equals to I into R 1. What is total current I? 1 ampere multiplied by R 1 10 ohm. So, V R 1 is equals to 10 volts. V R 2 I into R 2 is equals to I into 20 plus is equals to 20 volts. V R 3 is equals to I into R 3 is equals to I into 30. I is equals to 1 ampere into 30 ohm comes out to be 30 volts. V R 4 will be equals to again the total current multiplied by resistance R 1 is equals to 1 ampere multiplied by 40 ohm comes out to be 40 volts. Question, while tracing the circuit, if many, if my direction and the current direction is same, then voltage drop across any resistance, is it a positive or negative? Second question, if both the direction, my direction and the current direction is same, what will the polarity of the voltage drop? The answer, in the first case, if both the directions, my direction and the tracing direction and the current direction are same, voltage drop across any resistance will be negative. In the second case, if both the direction, that is my direction and the current direction is opposite, the drop will be treated as positive. Accounting, acknowledgement, thank you.