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Three problems on circuit analysis

Please solve problems 26.60, 26.62, and 26.64. See attached file for full problem description.

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SOLUTION

Problem 26.60
I
A B C
I1 (I - I1)

1 ohm 2 ohm
1 ohm
14 V D E
I2 (I - I1 - I2)
2 ohm (I1+I2) 1 ohm
I
F G H

Let us assume the currents as shown in the circuit diagram above.

We apply Kirchhoff's loop law to the loop ABGF starting from the negative terminal of the battery. We take voltage rise as positive and go around the loop in clockwise direction.

Voltage rise in battery - voltage drop in 1 ohm resistance - voltage drop in 2 ohm resistance = 0

14 - I1 - 2(I1+I2) = 0

Or 3I1 + 2I2 = 14 ...........(1)

We apply Kirchhoff's loop law to the loop BCED starting from point C.

- Voltage drop in 2 ohm resistance - voltage drop in 1 ohm resistance + voltage rise in 1 ohm resistance* = 0

[* because we are going against the direction of current]

- 2(I - I1) - I2 + I1 = 0

Or 2I - 3I1 + I2 ...

Solution Summary

These problems on circuit analysis have been solved step by step.

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