Kirchhoff’s Rules Problems and Solutions

 Problem#1

The batteries shown in the circuit in Fig. 1 have negligibly small internal resistances. Find the current through (a) the 30.0 Ω resistor; (b) the 20.0 Ω resistor; (c) the 10.0 V battery.

Fig.1

Answer:
Assume the unknown currents have the directions shown in Figure 2. We have used the junction rule to write the current through the 10.0 V battery as I₁ + I₂ . There are two unknowns, I₁ and I₂ , so we will need two equations. Three possible circuit loops are shown in the figure.

Fig.2

(a) Apply the loop rule to loop (1), going around the loop in the direction shown:

∑ε + ∑IR = 0
+10.0 V − (30.0 Ω)I₁ = 0
I₁ = 0.333 A.

(b) Apply the loop rule to loop (3):

∑ε + ∑IR = 0
+10.0 V − (20.0 Ω)I₂ − 5.00 V = 0
I₂ = 0.250 A

(c) the current through is

I = I₁ + I₂ = 0.333 A + 0.250 A = 0.583 A.

Problem#2
In the circuit shown in Fig. 3.  Find (a) the current in resistor R; (b) the resistance R; (c) the unknown emf ε (d) If the circuit is broken at point x, what is the current in resistor R?

Fig.3

Answer:

Fig.4

(a) Apply Kirchhoff’s junction rule to point a:

ΣI = 0
so I + 4.00 A − 6.00 A = 0

I = 2.00 A (in the direction shown in the diagram).

(b) Apply Kirchhoff’s loop rule to loop (1):

∑ε + ∑IR = 0
 −(6 00 A)(3.00 Ω) − (2.00 A)R + 28.0 V = 0
−18.0 V − (2.00 A)R + 28.0 V = 0
R = 10/2 = 5.00 Ω

(c) Apply Kirchhoff’s loop rule to loop (2):

∑ε + ∑IR = 0
−(6.00 A)(3.00 Ω) − (4.00 A)(6.00 Ω) + ε = 0
ε = 18.0 V + 24.0 V = 42.0 V.

(d) If the circuit is broken at point x there can be no current in the 6.00 Ω resistor. There is now only a single current path and we can apply the loop rule to this path.

Fig.5

∑ε + ∑IR = 0
+28.0 V – (3.00 Ω)I – (3.00 Ω)I – (5.00 Ω)I = 0
I = 28.0 V/8.00 Ω = 3.50 A

Problem#3
Find the emfs ε₁ and ε₂ in the circuit of Fig. 6, and find the potential difference of point b relative to point a.

Fig.6

Answer:
The circuit diagram is given in Figure 7.

Fig.7

The junction rule has been used to find the magnitude and direction of the current in the middle branch of the circuit. There are no remaining unknown currents.
The loop rule applied to loop (1) gives:

∑ε + ∑IR = 0
+20.0V − (1.00 A)(1.00 Ω) + (1.00 A)(4.00 Ω) + (1.00 A)(1.00 Ω) − ε₁ − (1.00 A)(6.00 Ω) = 0
ε₁ = 20.0 V − 1.00 V + 4.00 V +1.00 V − 6.00 V = 18.0 V

The loop rule applied to loop (2) gives:

∑ε + ∑IR = 0
+20.0 V − (1.00 A)(1.00 Ω) − (2.00 A)(1.00 Ω) − ε₂ − (2.00 A)(2.00 Ω) − (1.00 A)(6.00 Ω) = 0
ε₂ = 20.0 V − 1.00 V − 2.00 V − 4.00 V − 6.00 V = 7.0 V

Going from b to a along the lower branch,

V_b + (2.00 A)(2.00 Ω) + 7.0 V + (2.00 A)(1.00 Ω) = V_a ⋅
V_b − V_a = −13.0 V 

Point b is at 13.0 V lower potential than point a.

Problem#4
In the circuit shown in Fig. 8, find (a) the current in the 3.00 Ω resistor; (b) the unknown emfs ε₁ and ε₂ (c) the resistance R. Note that three currents are given.

Fig.8

Answer:

Fig.9

(a) Apply the junction rule to point a:

ΣI_a = 0
3.00 A + 5.00 A − I₃ = 0.
I₃ = 8.00 A

Apply the junction rule to point b:

ΣI_b = 0
2.00 A + I₄ − 3.00 A = 0
I₄ = 1.00 A

Apply the junction rule to point c:

ΣI_c = 0
I₃ − I₄ − I₅ = 0.
I₅ = I₃ − I₄
I₅ = 8.00 A −1.00 A = 7.00 A

(b) Apply the loop rule to loop (1):

∑ε + ∑IR = 0
ε₁ − (3.00 A)(4.00 Ω) − I₃(3.00 Ω) = 0.
ε₁ = 12.0 V + (8.00 A)(3.00 Ω) = 36.0 V

Apply the loop rule to loop (2):

∑ε + ∑IR = 0
ε₂ − (5.00 A)(6.00 Ω) − I₃(3.00 Ω) = 0
ε₂ = 30.0 V + (8.00 A)(3.00 Ω) = 54.0 V

(c) Apply the loop rule to loop (3):

∑ε + ∑IR = 0
−(2.00 A)R − ε₁ + ε₂ = 0
−(2.00 A)R – 36.0 V + 54.0 V = 0
R = 9.00 Ω

Share :