Using Kirchhoff's rules, find the following. (₁ = 70.7 V, ₂ = 60.9 V, and 3 = 78.7 V.) 4.00 kn www R₂ Step 1 b a E₁ 2.00 kn ww R₁ C E₂ R₂2 13.00 ΚΩ d Es (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. (₁ = 70.7 V, ₂ = 60.9 V, and 3 = 78.7 V.)
4.00 kn
www
R₂
b
a
E₁
So
2.00 kn
ww
R₁
1₁
so
Step 1
We assume the currents in the three branches of this circuit are directed as shown in the diagram below.
4.00 ΚΩ
d
ww
R3
(a) the current in each resistor shown in the figure above
(b) the potential difference between points c and f
a
12
C
2.00 ΚΩ
ww
R₁
If our assumed directions for the currents are correct, the calculated currents will have positive signs.
Solving for I, we have
E₂
R₂2 3.00 ΚΩ
We apply Kirchhoff's junction rule at point c to obtain
1₂ = 1₁ + 13.
[1]
R2 € 3.00 ΚΩ
We apply Kirchhoff's loop rule, going clockwise around loop abcfa to obtain
E₁-E₂-R₂I₂R₁1₁ = 0.
d
Es
Solving for I, we have
]) = 4
£₁ — &z - Ryb2 = (²₁ —-€²) - (22²) ¹2
R₁
1₁ = 5.1
xx10-3A)-([
[2]
Applying Kirchhoff's loop rule, going counterclockwise around loop edcfe, we have
E3-R313-₂-R₂I₂ = = 0.
´ E, - E2 - Ryb₂ = (B₁ −³2) - (223)³²
13 =
R3
x
10-³ A)-(C
²₂. [3]
Transcribed Image Text:Using Kirchhoff's rules, find the following. (₁ = 70.7 V, ₂ = 60.9 V, and 3 = 78.7 V.) 4.00 kn www R₂ b a E₁ So 2.00 kn ww R₁ 1₁ so Step 1 We assume the currents in the three branches of this circuit are directed as shown in the diagram below. 4.00 ΚΩ d ww R3 (a) the current in each resistor shown in the figure above (b) the potential difference between points c and f a 12 C 2.00 ΚΩ ww R₁ If our assumed directions for the currents are correct, the calculated currents will have positive signs. Solving for I, we have E₂ R₂2 3.00 ΚΩ We apply Kirchhoff's junction rule at point c to obtain 1₂ = 1₁ + 13. [1] R2 € 3.00 ΚΩ We apply Kirchhoff's loop rule, going clockwise around loop abcfa to obtain E₁-E₂-R₂I₂R₁1₁ = 0. d Es Solving for I, we have ]) = 4 £₁ — &z - Ryb2 = (²₁ —-€²) - (22²) ¹2 R₁ 1₁ = 5.1 xx10-3A)-([ [2] Applying Kirchhoff's loop rule, going counterclockwise around loop edcfe, we have E3-R313-₂-R₂I₂ = = 0. ´ E, - E2 - Ryb₂ = (B₁ −³2) - (223)³² 13 = R3 x 10-³ A)-(C ²₂. [3]
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