- Consider The Circuit In The Figure To The Right Here E 5 0 V E 10 0 V R 5 0 2 R 10 0 2 And R3 15 0 2 1 (156.69 KiB) Viewed 66 times
Consider the circuit in the figure to the right. Here, E₁ = 5.0 V, E₂ = 10.0 V, R₁ = 5.0 2, R₂ = 10.0 2 and R3 = 15.0 £2, 4. In this circuit diagram, draw arrows representing the loop and currents in each branch, make sure label each arrow and loop appropriately. Note: The direction of each loop and the direction of each current arrow that you are going to draw on the circuit above are arbitrary. Just assign voltage drops consistently and sum both voltage drops and currents algebraically and you will get the correct system equations. If the actual current is in the opposite direction from current arrow you choose, your answer for that current will be negative. The direction of any loop is even less important: The equation obtained from a counterclockwise loop is the same as that from a clockwise loop except for a negative sign in front of every term (i.e., an inconsequential change in overall sign of the equation because it equals zero). 5. Apply Kirchhoff's rules to the circuit above and write three equations only in terms of the circuit elements in the figure above (R₁, E₁, etc...) and the currents you assign at each junction. 1. 2. 3. 6. Solve the three equations that you wrote above for the values of the currents. Show your work. 4₁ = A R₂ 1₂ = A 13 = A