I don't understand from your description where points A and B are.
The two parallel resistors of 10 and 2 ohms are equivelant to 1 2/3 ohms in series. The two parallel resistors of 20 and 5 ohms are equivalent to 4 ohms in series.
Total resistance that the battery see is
3 + 1 2/3 + 4 + 6 ohms = 14 2/3
Use this and ohm's law to get the current in the series parts of the circuit.
A battery with voltage drop of 120V is connected in a ciruit containing both series and parallel. Moving in a clockwise direction. R1= 3 ohms there is a connection of point A to point B. In between there is a parallel circuit with R2= 10 ohms to R3=2ohms on the top. On the bottom there is another parallel circuit consisting of R4= 20 ohms and R5= 5 ohms. After point B is one more resistor R6= 6 ohms. Find the voltage drop and current across each resistor.
V=IR
I have been trying for hours to solve this problem but I just can't get the right answer. I can only figure out the information for R1(10A and 30V) and R6(10A and 60V). Can you help me with the rest?
2 answers
The current flowing through an electrical circuit is indirectly proportional to the resistance in the circuit. If a current of 8A flows through the circuit when the resistance is 30 ohms, how much current will flow through the circuit if the resistance is 14 ohms. show your answer to 3 significant figures.
1 answer