series-parallel circuit

Ohm's Law used in series-parallel found the total resistance? Rt= 2kΩ + { (12kΩ)(6kΩ)/(12kΩ+6kΩ)} Rt= 2kΩ + 4kΩ Rt= 6kΩ found the total current ? It= Vt/ Rt It= 54v / 6kΩ It= 9mA The current will be the same in the entire circuit but when it passes to the resistor in parallel it will be divided depending of the resistance value in parallel. found the total voltage? Vt= It × Rt V = 9mA × 6kΩ V = 54v found the total resistance? Rt= Vt/ It Rt= 54v / 9mA Rt=6kΩ If you notice the total current of the circuit is (It=9mA)this current pass completely for the first resistance( 2kΩ ) in series because the current will be the same level in series. The voltage drops when pass to the first resistor in series (voltage drops to 36v) How? If we multiply the current of the resistor in serie that is 9mA by 2kΩ wich is the first resisitor we will have the voltage of 18volt So the total voltage of 54 volt - 18 volt of the serostor in serie = 36volt VR1= IR1× R1= 18 volt VR1= 54v-18volt=( 36volt) pass the first resistor Now the voltage will be 36volt. It will pass thru the circuit instead of 54volt because the resistor in series cause dropping the voltage from 54 volt to 36volt. Now we know that the total current in a series circuit is the same always. In this case is 9mA. However, the total current of 9mA will be now dropping when it pass to the two resistors in parallel. First the total current of 9mA in a circuit will be dropping as always. It will go to the first parallel resistor with (R2 )12kΩ, so we use the formula of I = V / R using now ( 36 volt ) to be able to see the current that goes through the resistor of 12kΩ . I = V / R I R2= V / R2 I R2= 36V / 12kΩ the current in resistor 2 is (IR2 = 3mA) After that, we use the same formula to be able to found the current of the last resistor in parallel. I = V / R I R3= V / R3 I R3= 36V / 6kΩ the current in resistor 3 is ( I = 6mA) So when we sum the current of the resistors in parallel, the total current will be the same in the entire circuit. which is 3mA+ 6mA= 9mA all of this circuit is create by Jenko022 on via youtube

published 10 years ago