Electric current I=1 A passes through the bridge configuration shown in the figure below. The resistances of all 5 resistors are given: R=1.64 2, R2=11.33 2, R3=8.96 2, R4=8.14 and R5=4.96 2. Use Kirchhoff's rules to analyze the distribution of the currents inside this structure. We will denote the current running from point A to point B as I and the current running from point B to point C as 12. Then all other currents in different resistors can be expressed via I, I and I2. This way you will be able to formulate the appropriate system of equations to solve for the unknown currents I and I- R A B Rs R R3 R What is the electric current I?: 11= A. What is the electric current I2?: 12= A. Find now the potential difference (voltage) V=I R+12 R2 between points A and C. You can thus evaluate the equivalent resistance R=V/I of this bridge combination. What is R?: . R=