Exercise 26.30 Part A The 5.00-V battery in (Figure 1) is removed from the circu

Question

Exercise 26.30 Part A The 5.00-V battery in (Figure 1) is removed from the circuit and replaced by a 15.00-V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure Find the current in the upper branch. Express your answer with the appropriate units. Enter positive value if the current is to the left and negative value if the the current is to the right. Value Units 1= Value Submit My Answers Give Up Part B Find the current in the middle branch. Express your answer with the appropriate units. Enter positive value if the current is to the left and negative value if the the current is to the right. Figure 1 2.00( 1- Value Units 10.00 V a3.00 Submit My Answers Give Up 1.00 5.00 V Part C 4.00 Find the current in the lower branch Express your answer with the appropriate units. Enter positive value if the current is to the left and negative value if the the current is to the right. 10.00 1 Value Units

Explanation / Answer

Let the current in Upper Branch, = I1
Let the current in Middle Branch, = I2
Let the current in Lower Branch, = I3

I2 = I1 + I3 ----------1

I2*1 + I1 * 2.0 + 10.0 + I1*3.0 + I2 * 4.0 = 15.0
5*I2 + 5*I1 = 5.0
I2 + I1 = 1.0 ----------2

I2*1 + I3*10.0 + I2 * 4.0 = 15.0
5*I2 + 10*I3 = 15.0
I2 + 2*I3 = 3.0 --------3

Solving eq 1, 2 & 3
I1 = 0
I2 = 1
I3 = 1

Current in Upper Branch, = 0
Current in Middle Branch, = 1 A
Current in Lower Branch, = 1 A

Vab = - 10.0 - 1.0 + 15.0
Vab = 4.0V

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