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iris [78.8K]
3 years ago
10

A circuit is comprised of two adjacent square loops that share a side such that the top side of the bottom loop is also the bott

om side of the top loop. The bottom side of the bottom loop, from left to right, contains a 5.00 ⦠resistor and a 36.0 V battery, oriented such that the positive terminal is on the right and the negative is on the left. A current I3 runs from left to right. The top side of the top loop, from left to right, contains an 18.0 V battery, oriented such that the positive terminal is on the left and the negative is on the right, and a 5.00 ⦠resistor. A current I1 runs from right to left. The left side of the top loop contains an 8.00 ⦠resistor. The shared side, from left to right, contains an 11.0 ⦠resistor, a 12.0 V battery, oriented such that the positive terminal is on the right and the negative is on the left, and a 7.00 ⦠resistor. A current I2 runs from left to right. (a) Use Kirchhoff's rules to complete the equation for the upper loop. (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.) 30 V = (b) Use Kirchhoff's rules to complete the equation for the lower loop. (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.) â24 V = (c) Use Kirchhoff's rules to obtain an equation for the junction on the left side. (Use any variable or symbol stated above as necessary. All currents are given in amperes.) (d) Solve the junction equation for I3. I3 = (e) Using the equation found in part (d), eliminate I3 from the equation found in part (b). (Use any variable or symbol stated above as necessary. All currents are given in amperes. Do not enter any units.) 24 V = (f) Solve the equations found in part (a) and part (e) simultaneously for the two unknowns for I1 and I2, respectively. I1 = A I2 = A (g) Substitute the answers found in part (f) into the junction equation found in part (d), solving for I3. I3 = A (h) What is the significance of the negative answer for I2?

Physics
1 answer:
Airida [17]3 years ago
6 0

Answer:

a) 13I₁ + 18I₂ = 30v

b) 18I₁ - 5I₃ = -12v

c) I₁ = I₂ + I₃

d) I₃ = I₁ - I₂

e) 5I₁ - 23I₂ = 24v

f) I₁ =2.88A

  I₂ = -0.41A

g) I₃ = 3.29A

h) I₂ = 1.41A

The significance of the negative sign of I₂ is that the direction of the flow of current I₂ is opposite to the direction assumed.

Explanation:

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