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, Rtotal = 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:
Rtotal = 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, Rs = 16.0 Ω
resistors 1 and 2 are connected in parallel, Rp = 3.0 Ω
the equvalent resistance of two resistor connected in parallel is:
RP = R1R2/(R1 + R2)
3 = R1R2/(R1 + R2) (1)
the equvalent resistance of two resistor connected in parallel is:
RP = 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:
RP = R/4 = 4.5 Ω
the current through the battery is
i = V/Rp = 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.
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| Fig.1 |
Answer;
(a) the current in resistor 1
∑IR + ∑ε = 0
– i1R1 + ε2 = 0
i1 = ε2/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.
Vb – Va = – ε2 – ε3 = – 5.0 V – 4.0 V = – 9.0 V, or
Vb – Va = –i1R1 – ε3 = 0.05 A x 100 Ω – 4.0 V = –9.0 V, or
Vb – Va = – 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?
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| 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
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