Look at the diagram in procedure 1. Suppose the voltage across each battery is 2

Question

Look at the diagram in procedure 1.
Suppose the voltage across each battery is 29.4 V, and the resistance of each resistor is 23.1 ?.
NOTE: For this questions you will be allowed three submissions.

a) Find the current I2 (through R2).
I2 =  A

b) Suppose that wire AB breaks. Find the new current I2 (through R2).
I2 =  A

c) Suppose that, in the original circuit, the polarity of V1 is reversed, with all the values remaining unchanged. Find the new current I2(through R2).
I2 =  A

d) Suppose now that we start with the original circuit, but this time we add another branch identical to AF and CD, containing a resistorR = 23.1 ? and a battery V = 29.4 V, arranged with negative terminal toward the top.

Explanation / Answer

a) Applying KVL in left loop,

Va - i1R1 - 2i1R2 = 0

29.4 = i1 ( 23.1 + 2*23.1)

i1 = 0.424 A

i2 = 2i1 = 0.848 A

b) in this case i = 29.4 / (23.1 + 23.1) = 0.636 A

c) 29.4 = i1*23.1 + (i1-i3)23.1

46.2i1 - 23.1i3 = 29.4

in right loop,

i2*23.1 + (i3-i1)23.1 = 29.4

46.2i3 - 23.1i1 = 29.4

solving

i1 = 1.27A

i3 = 1.27 A

i2 = i1 - i2 = 0

d) Applying KVL in each loop,

29.4 - 23.1i1 - 29.4 - 23.1(i1-i2) = 0

46.2i1 = 23.1i2 ........(i)

29.4 - 23.1(i2 -i1) - 23.1*2(i2 -i1) = 0

69.3i2 - 69.3i1 = 29.4

putting i2 = 2i1

2*69.3i1 - 69.3i1 = 29.4

i1 = 0.423 A

i2 = 0.846 A

I2 = 2(i2 - i1) = 0.846 A

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