Kirchhoff’s Rules Problems and Solutions 2

 Problem#1

In the circuit shown in Fig. 1, find (a) the current in each branch and (b) the potential difference V_ab of point a relative to point b.

Fig.1

Answer:

Since the 10.0 V battery has the larger voltage, assume I₁ is to the left through the 10 V battery, I₂ is to the right through the 5 V battery, and I₃ is to the right through the 10 Ω resistor. Go around each loop in the counterclockwise direction.

Fig.2

(a) Upper loop:

∑ε + ∑IR = 0
10.0 V − (2.00 Ω + 3.00 Ω)I₁ − (1.00 Ω + 4.00 Ω)I₂ − 5.00 V = 0
This gives
5.0 V − (5.00 Ω)I₁ − (5.00Ω)I₂ = 0
I₁ + I₂ = 1.00 A

Lower loop:

∑ε + ∑IR = 0
5.00 V + (1.00 Ω + 4.00 Ω)I₂ − (10.0 Ω)I₃ = 0
This gives
5.00 V + (5.00 Ω)I₂ − (10.0 Ω)I₃ = 0
I₂ − 2I₃ = −1.00 A

Along with I₁ = I₂ + I₃ we can solve for the three currents and find:

I₁ = 0.800 A, I₂ = 0.200 A and I₃ = 0.600 A.

(b) the potential difference V_ab of point a relative to point b is

V_ab = −(0.200 A)(4.00Ω) − (0.800 A)(3.00 Ω) = −3.20 V

Problem#2
In the circuit shown in Fig. 3  the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15.0 V. (a) Find the emf ε of the battery. (b) What will the ammeter read when the switch is closed?

Fig.3

Answer:

(a) Find the current through the unknown battery using Ohm’s law. Then use the equivalent resistance of the circuit to find the emf of the battery.

The 30.0 Ω and 50.0 Ω resistors are in series, and hence have the same current. Using Ohm’s law
 I₅₀ = I₃₀ = (15.0 V/50.0 Ω) = 0.300 A

The potential drop across the 75.0 Ω resistor is the same as the potential drop across the 80.0 Ω series combination.

We can use this fact to find the current through the 75.0 Ω resistor using Ohm’s law:

V₇₅ = V₈₀ = (24 V/80.0 Ω) = 24.0 V
I₇₅ = (24.0 V)/(75.0 Ω) = 0.320 A

The current through the unknown battery is the sum of the two currents we just found:

 I_TOTAL = 0.300 A + 0.320 A = 0.620 A

The equivalent resistance of the resistors in parallel is

1/R_P = 1/(75.0 Ω) + 1/(80.0 Ω)
R_P = 38.7 Ω

The equivalent resistance “seen” by the battery is

 R_equiv = 20.0 Ω + 38.7 Ω = 58.7 Ω

Applying Ohm’s law to the battery gives

Total ε = R_equivI (58.7 ) x (0.620 A) = 36.4 V

(b) With the switch closed, the 25.0 V battery is connected across the 50.0 Ω resistor.
Ohm’s law gives

I = (25.0 V)/(50.0 Ω) = 0.500 A

The current through the 50.0- resistor, Ω and the rest of the circuit, depends on whether or not the switch is open.

Problem#3
In the circuit shown in Fig. 4 both batteries have insignificant internal resistance and the idealized ammeter reads 1.50 A in the direction shown. Find the emf ε of the battery. Is the polarity shown correct?

Fig.4

Answer:

Take a loop around the outside of the circuit, apply the junction rule at the upper junction, and then take a loop around the right side of the circuit.

The outside loop gives

75.0 V – (12.0 Ω)(1.50 A) – (48.0 Ω)I₄₈ = 0
so I₄₈ = 1.188 A

At a junction we have

1.50A = I_ε + 1.188 A,
I_ε = 0.313 A

A loop around the right part of the circuit gives

∑ε + ∑IR = 0
ε − (48 Ω)(1.188 A) + (15.0 Ω)(0.313 A) = 0

ε = 52.3 V with the polarity shown in the figure in the problem

The unknown battery has a smaller emf than the known one, so the current through it goes against its polarity.

Problem#4
In the circuit shown in Fig. 5, the 6.0 Ω resistor is consuming energy at a rate of 24 J/s when the current through it flows as shown. (a) Find the current through the ammeter A. (b) What are the polarity and emf ε of the battery, assuming it has negligible internal resistance?

Fig.5

Answer:
P = I²R so the power consumption of the 6.0 Ω resistor allows us to calculate the current through it. Unknown currents I₁, I₂ and I₃ are shown in Figure 6.

Fig.6

The junction rule says that

I₁ = I₂ + I₃.

In Figure 26.34 the two 20.0 Ω resistors in parallel have been replaced by their equivalent (10.0 Ω)

(a) the current through the ammeter A.
P = I²R gives
24 J/s = I²(6.0 Ω)
I = 2.0 A

The loop rule applied to loop (1) gives:

∑ε + ∑IR = 0
−(2.0 A)(3.0 Ω) + (2.0 A)(6.0 Ω) + 25 V – (10.0 Ω + 19.0 Ω + 1.0 Ω)I₂ = 0
I₂ = (25 V – 18 V)/20.0 Ω = 0.233 A
(b) I₃ = I₁ – I₂ = 2.0 A – 0.233 A = 1.77 A

The loop rule applied to loop (2) gives:

∑ε + ∑IR = 0
−(2.0 A)(3.0 Ω + 6.0 Ω) + 25 V + (1.77 A)(17 Ω) –  (1.77 A)(13 Ω) = 0
ε = 25 V – 18 V – 53.1 V = 46.1 V
The emf is 46.1 V.  

Share :