Solutions for Exercises 12 to 19 - TD5:

Current and Resistance

Table of Contents
• Exercise 12: Electrical Power
• Exercise 13: Resistors in Series and Parallel
• Exercise 14: Resistors in Series and Parallel
• Exercise 15: Resistors in Series and Parallel
• Exercise 16: Resistors in Series and Parallel
• Exercise 17: Resistors in Series and Parallel
• Exercise 18: Kirchhoff's Rules
• Exercise 19: Kirchhoff's Rules

Exercise 12: Electrical Power

Problem: Calculate the daily cost of operating a lamp that draws a current of 1.70 A from
a 110 V line. Assume the cost of energy from the power company is $0.0600/kWh.

Solution:

To solve this problem, we need to: 1. Calculate the electrical power consumed by the
lamp 2. Convert this power to energy consumed per day 3. Calculate the cost of this
energy

Step 1: Calculate the electrical power The electrical power is given by the formula: P =
V × I where: - P is the power in watts (W) - V is the voltage in volts (V) - I is the current
intensity in amperes (A)

With V = 110 V and I = 1.70 A: P = 110 V × 1.70 A = 187 W

Step 2: Calculate the energy consumed per day Energy is given by the formula: E = P ×
t where: - E is the energy in joules (J) or kilowatt-hours (kWh) - P is the power in watts (W)
- t is the time in seconds (s) or hours (h)

For a complete day (24 hours): E = 187 W × 24 h = 4,488 Wh = 4.488 kWh

Step 3: Calculate the daily cost Cost = E × rate where: - E is the energy in kWh - rate is
the cost per kWh
With E = 4.488 kWh and rate = $0.0600/kWh: Cost = 4.488 kWh × $0.0600/kWh = $0.26928

Final answer: The daily cost of operating the lamp is $0.27 (rounded to two decimal
places).

Exercise 13: Resistors in Series and Parallel

Problem: Consider the circuit shown in the figure, find: 1. The current in the 20Ω resistor
2. The potential difference between points a and b.

Solution:

To solve this problem, we need to analyze the circuit which contains: - A voltage source
of 25.0 V - Several resistors: 10.0 Ω (top), 10.0 Ω (horizontal), 5.00 Ω (left), 5.00 Ω (bottom)
and 20.0 Ω (right)

Step 1: Analyze the circuit structure The circuit presents a combination of series and
parallel connections: - A 10.0 Ω resistor in series with the voltage source - Three parallel
branches between points a and b: * A branch with a 5.00 Ω resistor (left) * A branch with
a 10.0 Ω resistor (middle) * A branch with 5.00 Ω and 20.0 Ω resistors in series (right)

Step 2: Calculate the equivalent resistance between points a and b For the three
parallel branches: - R₁ = 5.00 Ω (left branch) - R₂ = 10.0 Ω (middle branch) - R₃ = 5.00 Ω +
20.0 Ω = 25.0 Ω (right branch)

The equivalent resistance of the parallel branches is: 1/Req = 1/R₁ + 1/R₂ + 1/R₃ 1/Req =
1/5.00 + 1/10.0 + 1/25.0 1/Req = 0.2000 + 0.1000 + 0.0400 = 0.3400 Req = 1/0.3400 = 2.94 Ω

Step 3: Calculate the total current in the circuit The total resistance of the circuit is:
Rtot = 10.0 Ω + 2.94 Ω = 12.94 Ω

The total current is: I = V/Rtot = 25.0 V / 12.94 Ω = 1.93 A

Step 4: Calculate the potential difference between points a and b The voltage drop
across the 10.0 Ω resistor in series with the source is: V₁₀Ω = I × 10.0 Ω = 1.93 A × 10.0 Ω
= 19.3 V

The potential difference between points a and b is therefore: Vab = V - V₁₀Ω = 25.0 V - 19.3
V = 5.7 V

Step 5: Calculate the current in the 20.0 Ω resistor The 20.0 Ω resistor is in series with
a 5.00 Ω resistor in the right branch. The total resistance of this branch is 25.0 Ω.

The current in this branch is: I₂₀Ω = Vab / 25.0 Ω = 5.7 V / 25.0 Ω = 0.228 A
Final answer: 1. The current in the 20.0 Ω resistor is 0.228 A (or 228 mA) 2. The potential
difference between points a and b is 5.7 V

Exercise 14: Resistors in Series and Parallel

Problem: Three 100Ω resistors are connected as shown in the figure. The maximum
power that can safely be delivered to any one resistor is 25 W.

(a) What is the maximum voltage that can be applied to terminals a and b? (b) For the
voltage determined in part (a), what is the power delivered to each resistor? What is the
total power delivered?

Solution:

According to the figure, we have three 100Ω resistors in a configuration where one
resistor is in series with two resistors in parallel.

Step 1: Analyze the circuit structure - One 100Ω resistor is connected in series between
point a and the intermediate node - Two 100Ω resistors are connected in parallel
between this intermediate node and point b

Step 2: Calculate the equivalent resistance of the circuit For the two 100Ω resistors in
parallel: 1/Req_parallel = 1/100 + 1/100 = 2/100 Req_parallel = 100/2 = 50Ω

The total resistance of the circuit is: Rtot = 100Ω + 50Ω = 150Ω

Step 3: Determine the maximum voltage (part a) The maximum power for each
resistor is 25W.

For the series resistor (R1 = 100Ω): P = V²/R V = √(P×R) = √(25W × 100Ω) = 50V

Therefore, the maximum current in this resistor is: I = V/R = 50V/100Ω = 0.5A

This current also flows through the parallel resistors. For each parallel resistor (R2 = R3 =
100Ω), the current divides equally: I_parallel = I/2 = 0.5A/2 = 0.25A

The power in each parallel resistor is: P_parallel = I_parallel² × R = (0.25A)² × 100Ω =
6.25W

This power is below the 25W limit, so the series resistor is the limiting factor.

The maximum voltage across terminals a and b is: V_max = I × Rtot = 0.5A × 150Ω = 75V

Let's verify if this voltage respects the power limit for the parallel resistors: The voltage
across the parallel resistors is: V_parallel = I × Req_parallel = 0.5A × 50Ω = 25V
The power in each parallel resistor is: P_parallel = V_parallel²/R = (25V)²/100Ω = 6.25W

Therefore, the maximum voltage that can be applied to terminals a and b is 75V.

Step 4: Calculate the power delivered to each resistor (part b) For the maximum
voltage of 75V:

• Power in the series resistor (R1): P1 = I² × R1 = (0.5A)² × 100Ω = 25W

• Power in each parallel resistor (R2 and R3): P2 = P3 = I_parallel² × R = (0.25A)² ×

100Ω = 6.25W

• Total power delivered: Ptot = P1 + P2 + P3 = 25W + 6.25W + 6.25W = 37.5W

Final answer: (a) The maximum voltage that can be applied to terminals a and b is 75V.
(b) For this voltage: - The power delivered to the series resistor is 25W. - The power
delivered to each of the two parallel resistors is 6.25W. - The total power delivered is
37.5W.

Exercise 15: Resistors in Series and Parallel

Problem: The current in a circuit is tripled by connecting a 500Ω resistor in parallel with
the resistance of the circuit. Determine the resistance of the circuit in the absence of the
500Ω resistor.

Solution:

To solve this problem, we need to use the laws of electrical circuits and the formulas for
parallel resistors.

Step 1: Define the variables - Let R be the initial resistance of the circuit (unknown) - Let
I be the initial current in the circuit - Let I' be the current after adding the 500Ω resistor in
parallel (I' = 3I) - Let V be the voltage across the circuit (constant)

Step 2: Apply Ohm's law for the initial circuit According to Ohm's law: V = R × I
Therefore: I = V/R

Step 3: Calculate the equivalent resistance after adding the parallel resistor When
two resistors R and R' are in parallel, the equivalent resistance Req is given by: 1/Req = 1/
R + 1/R'

In our case: 1/Req = 1/R + 1/500

Step 4: Apply Ohm's law for the modified circuit After adding the 500Ω resistor in
parallel: V = Req × I' Therefore: I' = V/Req
Step 5: Use the relationship between I and I' We know that I' = 3I, so: V/Req = 3 × (V/R)

Simplifying: R/Req = 3

Substituting the expression for Req: R/(1/(1/R + 1/500)) = 3

Simplifying: R × (1/R + 1/500) = 3 1 + R/500 = 3 R/500 = 2 R = 1000Ω

Verification: If R = 1000Ω, then: 1/Req = 1/1000 + 1/500 = 1/1000 + 2/1000 = 3/1000 Req =
1000/3 = 333.33Ω

The ratio of currents is: I'/I = R/Req = 1000/333.33 = 3

This confirms that the current is indeed tripled.

Final answer: The resistance of the circuit in the absence of the 500Ω resistor is 1000Ω.

Exercise 16: Resistors in Series and Parallel

Problem: Calculate the power delivered to each resistor in the circuit shown in the
figure.

Solution:

The circuit contains: - A voltage source of 18.0 V - A 2.00 Ω resistor at the top - Two
resistors in parallel: 3.00 Ω and 1.00 Ω - A 4.00 Ω resistor at the bottom

Step 1: Analyze the circuit structure The circuit presents a combination of series and
parallel connections: - The 2.00 Ω resistor is in series with the rest of the circuit - The 3.00
Ω and 1.00 Ω resistors are in parallel with each other - The 4.00 Ω resistor is in series with
the parallel combination

Step 2: Calculate the equivalent resistance of the circuit For the two resistors in
parallel (3.00 Ω and 1.00 Ω): 1/Req_parallel = 1/3.00 + 1/1.00 = 0.333 + 1.000 = 1.333
Req_parallel = 1/1.333 = 0.75 Ω

The total resistance of the circuit is: Rtot = 2.00 Ω + 0.75 Ω + 4.00 Ω = 6.75 Ω

Step 3: Calculate the total current in the circuit I = V/Rtot = 18.0 V / 6.75 Ω = 2.67 A

Step 4: Calculate the currents in each branch - The current through the 2.00 Ω resistor
is the total current: I₁ = 2.67 A - The current through the 4.00 Ω resistor is the total
current: I₄ = 2.67 A - For the resistors in parallel, the current divides: * Voltage across the
parallel resistors: V_parallel = I × 0.75 Ω = 2.67 A × 0.75 Ω = 2.00 V * Current in the 3.00 Ω
resistor: I₃ = V_parallel / 3.00 Ω = 2.00 V / 3.00 Ω = 0.67 A * Current in the 1.00 Ω resistor:
I₁Ω = V_parallel / 1.00 Ω = 2.00 V / 1.00 Ω = 2.00 A

Step 5: Calculate the power delivered to each resistor The power is given by the
formula: P = I² × R

• Power in the 2.00 Ω resistor: P₂Ω = (2.67 A)² × 2.00 Ω = 7.13 A² × 2.00 Ω = 14.26 W

• Power in the 3.00 Ω resistor: P₃Ω = (0.67 A)² × 3.00 Ω = 0.45 A² × 3.00 Ω = 1.35 W

• Power in the 1.00 Ω resistor: P₁Ω = (2.00 A)² × 1.00 Ω = 4.00 A² × 1.00 Ω = 4.00 W

• Power in the 4.00 Ω resistor: P₄Ω = (2.67 A)² × 4.00 Ω = 7.13 A² × 4.00 Ω = 28.52 W

Verification: The total power delivered by the source is: Ptot = V × I = 18.0 V × 2.67 A =
48.06 W

The sum of the powers in the resistors is: Psum = 14.26 W + 1.35 W + 4.00 W + 28.52 W =
48.13 W

The slight difference is due to rounding in the calculations.

Final answer: - Power delivered to the 2.00 Ω resistor: 14.3 W - Power delivered to the
3.00 Ω resistor: 1.3 W - Power delivered to the 1.00 Ω resistor: 4.0 W - Power delivered to
the 4.00 Ω resistor: 28.5 W

Exercise 17: Resistors in Series and Parallel

Problem: For the circuits shown in figures (a) and (b), let R1=11Ω, R2=22Ω, and let the
battery have a terminal voltage of 33V.

(a) In the parallel circuit shown in figure (b), to which resistor is more power delivered?
(b) Verify that the sum of the power (I²R) delivered to each resistor equals the power
supplied by the battery (=I×ΔV). (c) In the series circuit (a), which resistor uses more
power? (d) Verify that the sum of the power (I²R) used by each resistor equals the power
supplied by the battery (=I×ΔV). (e) Which circuit configuration uses more power?

Solution:

Step 1: Analysis of the parallel circuit (figure b)

In a parallel circuit, the voltage is the same across each resistor: V = 33V for R1 and R2

The current in each branch is: - I₁ = V/R1 = 33V/11Ω = 3A - I₂ = V/R2 = 33V/22Ω = 1.5A

The total current is: I_total = I₁ + I₂ = 3A + 1.5A = 4.5A

The power delivered to each resistor is: - P₁ = I₁² × R1 = (3A)² × 11Ω = 9A² × 11Ω = 99W -
P₂ = I₂² × R2 = (1.5A)² × 22Ω = 2.25A² × 22Ω = 49.5W

Answer to question (a): More power is delivered to resistor R1 (99W compared to 49.5W
for R2).

Step 2: Verification of the total power in the parallel circuit

The total power delivered to the resistors is: P_resistors = P₁ + P₂ = 99W + 49.5W = 148.5W

The power supplied by the battery is: P_battery = I_total × V = 4.5A × 33V = 148.5W

Answer to question (b): The sum of the power delivered to each resistor (148.5W)
equals the power supplied by the battery (148.5W), which verifies the conservation of
energy.

Step 3: Analysis of the series circuit (figure a)

In a series circuit, the current is the same through each resistor: The total resistance is:
R_total = R1 + R2 = 11Ω + 22Ω = 33Ω The current is: I = V/R_total = 33V/33Ω = 1A

The power delivered to each resistor is: - P₁ = I² × R1 = (1A)² × 11Ω = 11W - P₂ = I² × R2 =

(1A)² × 22Ω = 22W

Answer to question (c): More power is used by resistor R2 (22W compared to 11W for
R1).

Step 4: Verification of the total power in the series circuit

The total power delivered to the resistors is: P_resistors = P₁ + P₂ = 11W + 22W = 33W

The power supplied by the battery is: P_battery = I × V = 1A × 33V = 33W

Answer to question (d): The sum of the power used by each resistor (33W) equals the
power supplied by the battery (33W), which verifies the conservation of energy.

Step 5: Comparison of the two configurations

• Total power in the parallel circuit: 148.5W

• Total power in the series circuit: 33W

Answer to question (e): The parallel configuration uses more power (148.5W compared
to 33W for the series configuration).

Explanation: In a parallel circuit, the equivalent resistance is lower than any of the
individual resistances, which allows a higher total current and therefore a greater total
power. In a series circuit, the resistances add up, which limits the current and therefore
the total power.

Exercise 18: Kirchhoff's Rules

Problem: The ammeter shown in this figure reads 2.00 A. Find I₁, I₂, and ε.

Solution:

To solve this problem, we will apply Kirchhoff's laws: - The junction rule (conservation of
current): the sum of currents entering a node equals the sum of currents leaving it. - The
loop rule (conservation of energy): the sum of potential differences around a closed loop
is zero.

Step 1: Identify the circuit elements The circuit contains: - A 15.0 V voltage source - A
voltage source ε (unknown) - A 7.00 Ω resistor - A 5.00 Ω resistor - A 2.00 Ω resistor - An
ammeter reading 2.00 A

Step 2: Apply the junction rule According to the figure, the ammeter reads 2.00 A.
According to the junction rule and the current directions indicated: I₁ - I₂ = 2.00 A
Therefore: I₁ = I₂ + 2.00 A

Step 3: Apply the loop rule We will apply the loop rule to two different loops:

Loop 1 (upper loop): Starting from the upper left corner and moving clockwise: -I₁ ×
7.00 Ω + 15.0 V - 5.00 Ω × 2.00 A = 0 -7.00 Ω × I₁ + 15.0 V - 10.0 V = 0 -7.00 Ω × I₁ + 5.0 V = 0
I₁ = 5.0 V / 7.00 Ω = 0.714 A

Loop 2 (lower loop): Starting from the lower left corner and moving clockwise: I₂ × 2.00
Ω - ε + 5.00 Ω × 2.00 A = 0 2.00 Ω × I₂ - ε + 10.0 V = 0 ε = 2.00 Ω × I₂ + 10.0 V

Step 4: Calculate I₂ We have determined that I₁ = 0.714 A. From the equation I₁ = I₂ + 2.00
A, we get: I₂ = I₁ - 2.00 A = 0.714 A - 2.00 A = -1.286 A

The negative sign indicates that I₂ flows in the direction opposite to that indicated in the
figure.

Step 5: Calculate ε Substituting the value of I₂ into the equation ε = 2.00 Ω × I₂ + 10.0 V:
ε = 2.00 Ω × (-1.286 A) + 10.0 V = -2.572 V + 10.0 V = 7.428 V ≈ 7.43 V

Verification: Let's verify that our results satisfy the junction rule: I₁ - I₂ = 0.714 A - (-1.286
A) = 0.714 A + 1.286 A = 2.00 A ✓

Final answer: - I₁ = 0.714 A (or 714 mA) - I₂ = -1.286 A (or -1.286 A, flowing in the direction
opposite to that indicated) - ε = 7.43 V
Exercise 19: Kirchhoff's Rules
Problem: 1- Determine the current in each branch of the circuit shown in this figure. 2-
The circuit is connected for 2.00 min. a) Find the energy delivered by each battery. b)
Find the energy delivered to each resistor. c) Identify the types of energy transformations
that occur in the operation of the circuit.

Solution:

The circuit contains: - Two voltage sources: 4.00 V and 12.0 V - Five resistors: 8.00 Ω, 5.00
Ω, 1.00 Ω (twice) and 3.00 Ω

Step 1: Define the currents in each branch Let's define the currents in the different
branches: - I₁: current in the branch containing the 8.00 Ω resistor - I₂: current in the
branch containing the 5.00 Ω resistor and a 1.00 Ω resistor - I₃: current in the branch
containing the 3.00 Ω resistor and a 1.00 Ω resistor

Step 2: Apply the junction rule At the upper node: I₁ = I₂ + I₃

Step 3: Apply the loop rule Let's apply the loop rule to two independent loops:

Loop 1 (left): Starting from the 4.00 V source and moving clockwise: 4.00 V - 8.00 Ω × I₁ -
5.00 Ω × I₂ - 1.00 Ω × I₂ = 0 4.00 V - 8.00 Ω × I₁ - 6.00 Ω × I₂ = 0 Equation 1: 8.00 Ω × I₁ +
6.00 Ω × I₂ = 4.00 V

Loop 2 (right): Starting from the 12.0 V source and moving clockwise: 12.0 V - 3.00 Ω × I₃
- 1.00 Ω × I₃ - 5.00 Ω × I₂ - 1.00 Ω × I₂ = 0 12.0 V - 4.00 Ω × I₃ - 6.00 Ω × I₂ = 0 Equation 2:
6.00 Ω × I₂ + 4.00 Ω × I₃ = 12.0 V

Step 4: Solve the system of equations We have three equations: - I₁ = I₂ + I₃ - 8.00 Ω × I₁

+ 6.00 Ω × I₂ = 4.00 V - 6.00 Ω × I₂ + 4.00 Ω × I₃ = 12.0 V

Substituting I₁ = I₂ + I₃ into the second equation: 8.00 Ω × (I₂ + I₃) + 6.00 Ω × I₂ = 4.00 V
8.00 Ω × I₂ + 8.00 Ω × I₃ + 6.00 Ω × I₂ = 4.00 V 14.0 Ω × I₂ + 8.00 Ω × I₃ = 4.00 V Equation
3: 14.0 Ω × I₂ + 8.00 Ω × I₃ = 4.00 V

Multiplying equation 3 by 0.75 to have the same coefficient for I₃ as in equation 2: 10.5 Ω
× I₂ + 6.00 Ω × I₃ = 3.00 V

Subtracting this equation from equation 2: 6.00 Ω × I₂ + 4.00 Ω × I₃ - (10.5 Ω × I₂ + 6.00 Ω

× I₃) = 12.0 V - 3.00 V -4.50 Ω × I₂ - 2.00 Ω × I₃ = 9.00 V Equation 4: 4.50 Ω × I₂ + 2.00 Ω ×
I₃ = -9.00 V

Multiplying equation 4 by 4 to have the same coefficient for I₃ as in equation 3: 18.0 Ω ×

I₂ + 8.00 Ω × I₃ = -36.0 V
Subtracting this equation from equation 3: 14.0 Ω × I₂ + 8.00 Ω × I₃ - (18.0 Ω × I₂ + 8.00 Ω
× I₃) = 4.00 V - (-36.0 V) -4.00 Ω × I₂ = 40.0 V I₂ = -10.0 A

Substituting this value into equation 3: 14.0 Ω × (-10.0 A) + 8.00 Ω × I₃ = 4.00 V -140 V +
8.00 Ω × I₃ = 4.00 V 8.00 Ω × I₃ = 144 V I₃ = 18.0 A

Calculating I₁: I₁ = I₂ + I₃ = -10.0 A + 18.0 A = 8.00 A

Step 5: Calculate the energy delivered by each battery The connection time is 2.00
min = 120 s.

For the 4.00 V battery: E₁ = P × t = V × I₁ × t = 4.00 V × 8.00 A × 120 s = 3840 J = 3.84 kJ

For the 12.0 V battery: E₂ = P × t = V × I₃ × t = 12.0 V × 18.0 A × 120 s = 25920 J = 25.9 kJ

Step 6: Calculate the energy delivered to each resistor For the 8.00 Ω resistor: E₈Ω = P
× t = I₁² × R × t = (8.00 A)² × 8.00 Ω × 120 s = 6144 J = 6.14 kJ

For the 5.00 Ω resistor: E₅Ω = P × t = I₂² × R × t = (-10.0 A)² × 5.00 Ω × 120 s = 6000 J =
6.00 kJ

For the first 1.00 Ω resistor (in the I₂ branch): E₁Ω₁ = P × t = I₂² × R × t = (-10.0 A)² × 1.00
Ω × 120 s = 1200 J = 1.20 kJ

For the 3.00 Ω resistor: E₃Ω = P × t = I₃² × R × t = (18.0 A)² × 3.00 Ω × 120 s = 11664 J =
11.7 kJ

For the second 1.00 Ω resistor (in the I₃ branch): E₁Ω₂ = P × t = I₃² × R × t = (18.0 A)² ×
1.00 Ω × 120 s = 3888 J = 3.89 kJ

Total energy delivered to the resistors: Etot = 6.14 kJ + 6.00 kJ + 1.20 kJ + 11.7 kJ + 3.89
kJ = 28.93 kJ

Step 7: Identify the types of energy transformations In this circuit, the following
energy transformations occur:

1. Transformation of chemical energy into electrical energy in the batteries:

2. The 4.00 V battery transforms 3.84 kJ of chemical energy into electrical energy

3. The 12.0 V battery transforms 25.9 kJ of chemical energy into electrical energy

4. Transformation of electrical energy into thermal energy in the resistors:

5. The resistors transform a total of 28.93 kJ of electrical energy into thermal energy
(heat)

6. Conservation of energy:
 7. The total energy supplied by the batteries (3.84 kJ + 25.9 kJ = 29.74 kJ) is
approximately equal to the total energy dissipated by the resistors (28.93 kJ), the
difference being due to rounding in the calculations.

Final answer: 1- The currents in each branch are: - I₁ = 8.00 A (in the branch with the 8.00
Ω resistor) - I₂ = -10.0 A (in the branch with the 5.00 Ω and 1.00 Ω resistors) - I₃ = 18.0 A (in
the branch with the 3.00 Ω and 1.00 Ω resistors)

2- For a connection time of 2.00 min: a) The energy delivered by each battery is: - 4.00 V
battery: 3.84 kJ - 12.0 V battery: 25.9 kJ

b) The energy delivered to each resistor is: - 8.00 Ω resistor: 6.14 kJ - 5.00 Ω resistor: 6.00
kJ - First 1.00 Ω resistor: 1.20 kJ - 3.00 Ω resistor: 11.7 kJ - Second 1.00 Ω resistor: 3.89 kJ

c) The types of energy transformations are: - Transformation of chemical energy into

electrical energy in the batteries - Transformation of electrical energy into thermal
energy in the resistors - Conservation of total energy in the circuit