Multi-loop Circuits (EMF, Internal Resistance, and Terminal Voltage of Batteries) Problems and Solutions

 Problem #1

A total resistance of 3.00 Ω is to be produced by connecting an unknown resistance to a 12.0 Ω resistance. (a) What must be the value of the unknown resistance, and (b) should it be connected in series or in parallel?

Answer;
Known:
Resistance, R_total = 3.00 Ω and R₁ = 12.0 Ω

The total resistance is smaller than 12 Ω, therefore the second resistor is in parallel with the first resistor:

R_total = R₁R₂/(R₁ + R₂)

3 Ω = 12 ΩR₂/(12 Ω + R₂)

R₂ = 4.00 Ω

Problem #2
When resistors 1 and 2 are connected in series, the equivalent resistance is 16.0 Ω. When they are connected in parallel, the equivalent resistance is 3.0 Ω.What are (a) the smaller resistance and (b) the larger resistance of these two resistors?

Answer;
resistors 1 and 2 are connected in series, R_s = 16.0 Ω
resistors 1 and 2 are connected in parallel, R_p = 3.0 Ω

the equvalent resistance of two resistor connected in parallel is:

R_P = R₁R₂/(R₁ + R₂)      

3 = R₁R₂/(R₁ + R₂)              (1)

the equvalent resistance of two resistor connected in parallel is:

R_P = R₁ + R₂

R₁ = 16 – R₂                         (2)

Subtitute from (2) into (1) to eliminate R₁, so

3 = [16 – R₂]R₂/(16 – R₂ + R₂)

3 x 16 = 16R₂ – R₂²

R₂² – 16R² + 48 = 0

R₂ = 16 ± [16² – 4 x 48]/2

R₂ = 8.00Ω ± 4.0 Ω

(a) so we have two values for the resistor, so we can take any one of them because the other resistor will have the same values, so:

R₂ = 8.00Ω – 4.0 Ω = 4.0 Ω

(b) and R₁ = 16 – R₂ = 16.0 Ω – 4.0 Ω = 12.0 Ω

Problem #3
Four 18.0 Ω resistors are connected in parallel across a 25.0 V ideal battery. What is the current through the battery?

Answer;
Known:
resistance R = 18.0 Ω
electrical voltage, V = 25.0 V

the equvalent resistance of four two resistor connected in parallel is:

R_P = R/4 = 4.5 Ω

the current through the battery is

i = V/R_p = 25.0 V/4.5 Ω = 5.56 A

Problem #4
In Fig. 01, R₁ = 100 Ω, R₂ = 50 Ω, and the ideal batteries have emfs ε₁ = 6.0 V, ε₂ = 5.0 V, and ε₃ = 4.0 V. Find (a) the current in resistor 1, (b) the current in resistor 2, and (c) the potential difference between points a and b.

Fig.1

Answer;
(a) the current in resistor 1

∑IR + ∑ε = 0

– i₁R₁ + ε₂ = 0

i₁ = ε_2­/R₁ = 5.0 V/100 Ω = 0.05 A

(b) the current in resistor 2

∑IR + ∑ε = 0

– i₂R₂ + ε₁ – ε₂ – ε₃ = 0

i₂ = (ε₁ – ε₂ – ε₃)/R₂

   = (6.0 V – 5.0 V – 4.0 V)/50 Ω

i₂ = –0.06 A

(c) the potential difference between points a and b.

V_b – V_a = – ε₂ – ε₃ = – 5.0 V – 4.0 V = – 9.0 V, or

V_b – V_a = –i₁R₁ – ε₃ = 0.05 A x 100 Ω – 4.0 V = –9.0 V, or

V_b – V_a = – ε₁ – i₂R₂ = –6.0 V – (0.06 A x 50 Ω) = –9.0 V, or

Problem #5
Side flash. Figure 2 indicates one reason no one should stand under a tree during a lightning storm. If lightning comes down the side of the tree, a portion can jump over to the person, especially if the current on the tree reaches a dry region on the bark and thereafter must travel through air to reach the ground. In the figure, part of the lightning jumps through distance d in air and then travels through the person (who has negligible resistance relative to that of air because of the highly conducting salty fluids within the body).The rest of the current travels through air alongside the tree, for a distance h. If d/h = 0.400 and the total current is I = 5000 A, what is the current through the person?

Fig.2

Answer;

Known:
d/h = 0.400
total current, I = 5000 A

Since the potential differences across the two aths are the same,

V₁ = V₂

With V₁ for the left path, and V₂ for the right path, We have

i₁R₁ = i₂R₂,

Where i = i₁ + i₂ = 5000 A.

With R = ρL/A, the above equation can be rewritten as

i₁d = i₂h

i₂ = i₁(d/h) = (0.400)i₁, then

i₁ + (0.400)i₁ = 5000 A

we get, i₁ = 5000 A/1.400 = 3571.42 A

and i₂ = i ─ i₁ = 5000 A ─ 3571 A = 1428.57 A.

Thus, the current through the person is i₁ = 3571 A ≈ 3.6 kA