I can\'t get past the first part of this problem with thenumbers I was given. In

Question

I can't get past the first part of this problem with thenumbers I was given. In the figure shown in the first part ofthe solution, I was given 1=4 V,2= 4 V, R1 = 4 , R2 =4 , and R3 = 3 . Both batteries areideal. What is the rate at which energy is dissipated in eachresistor? There are a lot of parts to the problem and I don't know whereI went wrong, either setting up kirchhoffs law or solving for thecurrent? I can't get past the first part of this problem with thenumbers I was given. In the figure shown in the first part ofthe solution, I was given 1=4 V,2= 4 V, R1 = 4 , R2 =4 , and R3 = 3 . Both batteries areideal. What is the rate at which energy is dissipated in eachresistor? There are a lot of parts to the problem and I don't know whereI went wrong, either setting up kirchhoffs law or solving for thecurrent?

Explanation / Answer

Assume current i1 flows through the left loop with directionclockwise. Assume current i2 flows through the right loop with directionanti-clockwise so i1+i2 will flow through resistor R3 with directiondownwards. Take KVL to the left loop E1-R1*i1-R3*(i1+i2) = 0 4-4i1-3i1-3i2=0 7i1+3i2 = 4 --------(1) Take KVL to the right loop E2-R2*i2-R3*(i1+i2) = 0 4-4i2-3i1-3i2 = 0 -3i1-7i2 = -4 3i1+7i2 = 4----------(2) Equations right hand sides are equal So the left hand sides can be equal 7i1+3i2 = 3i1+7i2 4i1 = 4i2 i1=i2 So equation (1) becomes 7i1+3i1 = 4 i1 =0.4 A i2 = 0.4 A Now let us calculate the energy dissipated by eachresistor. Formula to use is I^2*R Energy in R1 = 0.4^2*4 = 0.64 W Energy in R2 will be the same as the current is same andresistor also same. 0.64W Energy in R3 = 0.8^2*3 = 1.92 W

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