Last Problem PLEASE HELP Me

For I1 you would use r1 and r3. If you look at the arrows for the current on either side, you can see that the current comes from either E1 or E2, flows through r1 or r2 respectively, and both currents flow through r3. Does that help?

 

Okay, my apologies, I didn't read the problem correctly. You should just use the resistance of r1--it is just looking for the flow of current across that resistor.

 

Kirchoffs Laws. you have two loops e1 and e2 the voltage drop accross each loop must sum to zero That means for left Loop E1 the voltage accros R1 and Voltage accross R3 must sum to E1. Voltage accross R1 is R1 x I1 Voltage accross R3 is R3 x (I1 + I2) THAT is the TRICK the Current in R3 is the sum of two currents I1 and I2. doing same fro right loop we get E2 = I2xR2 + (I1+I2)x R3 and from left loop E1 = I1xR1 +(I1+I2)x R3 simplify and group terms..E1 = I1x(R1+R3) + I2xR3 and E2 = I2x(R2+R3) + I1xR3 use values of r1 r2 and r3 E1 = I1x2.4 + I2x 4.6 and E2 = 5.6 x I2 + 1 x I1 This is two equations in two unkowns BUT he already told us e1 = 11 and I = to I1 + I2 = 4.11 so E1 = I1x2.4 +I2x4.6 so I1 = 4.11 - I2 so e1 = 11 = 2.4x4.11-2.4x I2 + 4.6 x I2----- 11 = 9.86 + 2.2 x I2 ..... I2 = 0.518 Amps

 

It is solved already We just worked at getting I2 = .518 Now go back to second loop eequation and sustitute what we have so far E2 = .518x(R2) + (I1+I2)xR3 But he gave you I = I3 = I1 + I2 = 11 so E2 = .518x4.6 + 11 x 1 or E2 = 13.383

 

What the problem calls I you are calling I3, so lets re-write E1 = I1R1 + I3R3 E2 = I2R2 +I3R3 He gave us I = 4.11 and E1 = 11V 11V = I1x1.4 + 4.11x 1 so I1 = (11-4.11)/1.4 so I1 = 4.72 A so I3 = 11 = I1 + I2 so 11 = 4.72 + I2 so I2 = 6.28 sorry I screwed up last try

 

Dis regard, I1 = 4.72 I2 = 6.28 so E2 = i2xr2 +i3xr3 6.28x4.6 + 11 x 1 = 39.9