Question

Question about resistors in a circuit. See picture

Answer 1

Calculate the current in the R5 = 8.56 Ω resistor.


Calculate the current in the R6 = 4.33 Ω resistor.
Calculate the current in the R3 = 15.7 Ω resistor.


Calculate the current in the R4 = 12.8 Ω resistor.

Answer 2

The current through R3 and R4 will be the same and so will the current through R5 and R6....since those pairs of resistors are in series. You can start by combining R3 and R4 into one resistor of 28.5 ohms and R5 and R6 into another resistor of 12.89 ohms to make things simpler.

Next, calculate the voltage across the middle parallel branch:\[V = IR = (14.4 \Omega )(0.726A)=10.45V\]Now you can find the current through R2:\[I_{R2}=\frac{V}{R}=\frac{10.45V}{17.4 \Omega}=0.601A\]
You know the voltage drop across the R1/R2 branch must be the same as that across R5/R6 since those two branches are in parallel, so you can find the current through R5/R6 by:\[I_{R5+R6}=\frac{10.45V}{12.89 \Omega}=0.811A\]
Now we just need to find the current on the left side (R3/R4). Since it's in series with the other 2 branches, it's going to be the sum of the parallel branch currents:\[I_{R3+R4}=0.811A+0.601A+0.726A=2.14A\]