Using Kirchhoff's rules, find the following. (E, = 70.9 V, E, = 61.7 V, and E, = 78.1 V.) 4.00 kN R 3.00 kM 2.00 kN R1 (a) the current in each resistor shown in the figure above (b) the potential difference between points c and f

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Using Kirchhoff's rules, find the following. (E, = 70.9 V, Ɛ, = 61.7 V, and ɛ, = 78.1 V.)
4.00 kN
E,
R.
R,S 3.00 kN
2.00 kN
R1
(a) the current in each resistor shown in the figure above
(b) the potential difference between points c and f
Step 1
We assume the currents in the three branches of this circuit are directed as shown in the diagram below.
4.00 k2
R3
R23.00 k.
2.00 k2
f
R1
If our assumed directions for the currents are correct, the calculated currents will have positive signs.
We apply Kirchhoff's junction rule at point c to obtain
I2 = I1 + I3.
[1]
We apply Kirchhoff's loop rule, going clockwise around loop abcfa to obtain
E, - E2 - R2I2 - R¿l1 = 0.
Solving for I,, we have
E, - E, - R2 - () -).
I =
R1
so
4 = (C
10-3 A) - (O )2: (2)
Applying Kirchhoff's loop rule, going counterclockwise around loop edcfe, we have
Ez - R3l3 - E2 - R2I2 = 0.
Solving for Ia, we have
8,- E, - Rzlz - ()·
I3 =
R3
so
Iz =
J- 10-A) - (C
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Transcribed Image Text:Using Kirchhoff's rules, find the following. (E, = 70.9 V, Ɛ, = 61.7 V, and ɛ, = 78.1 V.) 4.00 kN E, R. R,S 3.00 kN 2.00 kN R1 (a) the current in each resistor shown in the figure above (b) the potential difference between points c and f Step 1 We assume the currents in the three branches of this circuit are directed as shown in the diagram below. 4.00 k2 R3 R23.00 k. 2.00 k2 f R1 If our assumed directions for the currents are correct, the calculated currents will have positive signs. We apply Kirchhoff's junction rule at point c to obtain I2 = I1 + I3. [1] We apply Kirchhoff's loop rule, going clockwise around loop abcfa to obtain E, - E2 - R2I2 - R¿l1 = 0. Solving for I,, we have E, - E, - R2 - () -). I = R1 so 4 = (C 10-3 A) - (O )2: (2) Applying Kirchhoff's loop rule, going counterclockwise around loop edcfe, we have Ez - R3l3 - E2 - R2I2 = 0. Solving for Ia, we have 8,- E, - Rzlz - ()· I3 = R3 so Iz = J- 10-A) - (C Submit Skip (you cannot come back)
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