On the center BE branch, I read a ten ohm resistor in series with Ri, and Ri is in parallel with a ten volt source. If that is so, in the right loop, working clockwise, from F. The loop currents are IRight (IR) and Ileft.I put a + on the top of the ten ohm resistor.
10+10+10(Ir-Il) -5IR=0 (sum of voltage drops is zero, where Ir is the loop current.
Now on the left loop, starting at E.
10-27IL-10(Ir-Il)=0
those are the loop equations. Solve for Ir, Il.
The current through the 10 ohm resistor is Ir-Il.
Find the current in the 10.0 ohm resistor in the drawing (V1 = 21.0 V and R1 = 27.0 ohms).
I will attempt to explain the diagram again.
The drawing shows 2 parallel lines labelled as follows:
A------------B-------------C
D------------E-------------F
A and D are connected.
C and F are connected.
B and E are connected which represents an internal resistance.
The first resister R1=27.0 ohm is between A and B.
The second resister= 5.0 ohm is between B and C.
10.0v is introduced between D and E with the positive terminal in the left and the negative terminal on the right.
V1=21.0v is introduced between E and F with the positive terminal on the left and the negative terminal on the right.
The internal resistance has a resistor located closer to B whose value= 10.0 ohm.
10.0v is introduced ot the internal resistance located closer to point E. Its positive terminal is up and the negative terminal is on the bottom.
This question deals with kirchhoff's rules.
The current is moving form the bottom to the top.
1 answer
1 answer