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Resistance problems measuring current in a circuit

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See attached files.

1. Find the current in each resistor in attachment #1 which is #27 on the page and it is circled.

2. Find a). the equivalent resistance of the circuit in the second Attachment or #47 on page and it is circle as well,
b). each current in the circuit,
c). the potential difference across each resistor, and
d). the power dissipated by each resistor.

3. The student engineer of a campus radio station wishes to verify the effectiveness of the lightning rod on the antenna mast, this is attachment #3 and it says Figure P18.55. The unknown resistance Rx is between points c and e. Point e is a "true ground" but is inaccessible for direct measurement since this stratum is several meters below the earth's surface. Two identical rods are driven into the ground at a and b, introducing an unknown resistance Ry. The procedure is as follows: measure resistance R1 between points a and b, then connect a and b with a heavy conducting wire and measure resistance R2 between points a and c.
a). Derive a formula for Rx in terms of the observable resistances R1 and R2.
b). a satisfactory ground resistance would be Rx<2.0Ohms. Is the grounding of the station adequate if measurements give R1=13ohms and R2=6.0ohms?

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with detailed calculations and explanations, the problems are solved.

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See the attached file.

1.) Let the current in 30 ohm res. be i1 (left to right), in 5 ohm i2 (left to right) and in 20 ohm i3 (right to left).

Now by KCL (Kirchhoff's current law) at any one of two junctions:

i1 + i2 - i3 = 0
=>i3 = i1 + i2 ......eqn(1)

by KVL in biggest closed loop (20 V battery, 30 ohm and 20 ohm res.):
clockwise positive:
summation(i*R) = summation(emf)
30 * i1 + 20 * (i1+i2) = 20
=>50.i1 + 20.i2 = 20
=> 5.i1 + 2.i2 = 2 ......eqn(2)

for lower closed loop (10 V battery and 5 and ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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