Problem #1
In Fig. 1, R1 = 6.00 Ω, R2 = 18.0 Ω, and the ideal battery has emf ε = 12.0 V. What are the (a) size and (b) direction (left or right) of current i1? (c) How much energy is dissipated by all four resistors in 1.00 min? |
| Fig.1 |
Answer;
Known:
resistance R1 = 6.00 Ω,
resistance R2 = 18.0 Ω
emf ε = 12.0 V
time, t = 1.00 min = 60.0 s
(a) the parallel set of three identical R2 = 18 Ω resistors reduce to R = 6.0 Ω, which is now in series with the R1 = 6.0 Ω resistor at the top right, so that the total resistive load across the battery is
R’ = R1 + R = 12 Ω.
Thus, the current through R’ is
(12 V)/R’ = 1.0 A, which is the current through R.
By symmerty, we see one─third of that passes through any one of those 18 Ω resistors:
Therefore i1 = 0.33 A
(b) the direction of i1 is clearly rightward
(c) we use P = i2R’ = (1.0 A)2(12 Ω) = 12 W.
Thus in 60 s, the energy dissipated is
U = Pt = (12 W)(60 s) = 720 J
Problem #2
In Fig. 2, the ideal batteries have emfs ε1 = 10 V and ε2 = 0.500ε1, and the resistances are each 4.00 Ω . What is the current in (a) resistance 2 and (b) resistance 3?
Known:
resistance R1 = 6.00 Ω,
resistance R2 = 18.0 Ω
emf ε = 12.0 V
time, t = 1.00 min = 60.0 s
(a) the parallel set of three identical R2 = 18 Ω resistors reduce to R = 6.0 Ω, which is now in series with the R1 = 6.0 Ω resistor at the top right, so that the total resistive load across the battery is
R’ = R1 + R = 12 Ω.
Thus, the current through R’ is
(12 V)/R’ = 1.0 A, which is the current through R.
By symmerty, we see one─third of that passes through any one of those 18 Ω resistors:
Therefore i1 = 0.33 A
(b) the direction of i1 is clearly rightward
(c) we use P = i2R’ = (1.0 A)2(12 Ω) = 12 W.
Thus in 60 s, the energy dissipated is
U = Pt = (12 W)(60 s) = 720 J
Problem #2
In Fig. 2, the ideal batteries have emfs ε1 = 10 V and ε2 = 0.500ε1, and the resistances are each 4.00 Ω . What is the current in (a) resistance 2 and (b) resistance 3?
 |
| Fig.2 |
Known:
emfs ε1 = 10 V and ε2 = 0.500ε1
resistances, R1 = R2 = R3 = 4.00 Ω
Imagine the current flowing out of each emf:
Out of ε1 clockwise through R1, and out of ε2 counterclockwise through R2. Then the current through R3 must be just the sum of i1 and i2:
i3 = i1 + i2
we have 3 loops to apply kirchoff’s the two loops which flows through R3, but you can choose any 2 loops:
∑ε + ∑(iR) = 0
ε1 ─ i1R1 ─ i3R3 = 0
LOOP 1: 10 V ─ i1(4 Ω) ─ i3(4 Ω) = 0 or i1 + i3 = 2.5 (*)
LOOP 2: 5 V ─ i2(4 Ω) ─ i3(4 Ω) = 0 or i2 + i3 = 1.25 (**)
Because i3 = i1 + i2, then (*) and (**) to be
2i1 + i2 = 2.5 or i2 = 2.5 ─ 2i1 (***)
Subtitute from (***) into (**) so
i1 + 2(2.5 ─ 2i1) = 1.25
3i1 = 3.75
i1 = 1.25 A and i2 = 2.5 ─ 2i1 = 0 A
Problem #3
In Fig. 3, the current in resistance 6 is i6 = 1.40 A and the resistances are R1 = R2 = R3 = 2.00 Ω, R4 = 16.0 Ω, R5 = 8.00 Ω, and R6 = 4.00 Ω. What is the emf of the ideal battery?
emfs ε1 = 10 V and ε2 = 0.500ε1
resistances, R1 = R2 = R3 = 4.00 Ω
Imagine the current flowing out of each emf:
Out of ε1 clockwise through R1, and out of ε2 counterclockwise through R2. Then the current through R3 must be just the sum of i1 and i2:
i3 = i1 + i2
we have 3 loops to apply kirchoff’s the two loops which flows through R3, but you can choose any 2 loops:
∑ε + ∑(iR) = 0
ε1 ─ i1R1 ─ i3R3 = 0
LOOP 1: 10 V ─ i1(4 Ω) ─ i3(4 Ω) = 0 or i1 + i3 = 2.5 (*)
LOOP 2: 5 V ─ i2(4 Ω) ─ i3(4 Ω) = 0 or i2 + i3 = 1.25 (**)
Because i3 = i1 + i2, then (*) and (**) to be
2i1 + i2 = 2.5 or i2 = 2.5 ─ 2i1 (***)
Subtitute from (***) into (**) so
i1 + 2(2.5 ─ 2i1) = 1.25
3i1 = 3.75
i1 = 1.25 A and i2 = 2.5 ─ 2i1 = 0 A
Problem #3
In Fig. 3, the current in resistance 6 is i6 = 1.40 A and the resistances are R1 = R2 = R3 = 2.00 Ω, R4 = 16.0 Ω, R5 = 8.00 Ω, and R6 = 4.00 Ω. What is the emf of the ideal battery?
 |
| Fig.3 |
Known:
Firs, we note in V4, that the voltage across R4 is equal to
i4 = V4/R4 = 16.8 V/16.0 Ω = 1.05 A
By the junction rule, the current in R2 is
i2 = i4 + i6 = 1.05 A + 1.40 A = 2.45 A
So its voltage is
V2 = (2.00 Ω)(2.45 A) = 4.90 V
The loop rule tells us the voltage across R3 is
V3 = V2 + V4 = 21.7 V
implying that the current through it is i3 = V3/R3 = 10.85 A
The junction rule now gives the current in R1 as i1 = i2 + i3 = 2.45 A + 10.85 A = 13.3 A
Implying that the voltage across it is
V1 = (13.3 A)(2.00 Ω) = 26.6 V.
Therefore, by the loop rule,
ε = V1 + V3 = 26.6 V + 21.7 V = 48.3 V
Problem #4
The resistances in Figs. 6a and b are all 6.0 Ω, and the batteries are ideal 12 V batteries. (a) When switch S in Fig. 4a is closed, what is the change in the electric potential V1 across resistor 1, or does V1 remain the same? (b) When switch S in Fig. 4b is closed, what is the change in V1 across resistor 1, or does V1 remain the same?
Firs, we note in V4, that the voltage across R4 is equal to
i4 = V4/R4 = 16.8 V/16.0 Ω = 1.05 A
By the junction rule, the current in R2 is
i2 = i4 + i6 = 1.05 A + 1.40 A = 2.45 A
So its voltage is
V2 = (2.00 Ω)(2.45 A) = 4.90 V
The loop rule tells us the voltage across R3 is
V3 = V2 + V4 = 21.7 V
implying that the current through it is i3 = V3/R3 = 10.85 A
The junction rule now gives the current in R1 as i1 = i2 + i3 = 2.45 A + 10.85 A = 13.3 A
Implying that the voltage across it is
V1 = (13.3 A)(2.00 Ω) = 26.6 V.
Therefore, by the loop rule,
ε = V1 + V3 = 26.6 V + 21.7 V = 48.3 V
Problem #4
The resistances in Figs. 6a and b are all 6.0 Ω, and the batteries are ideal 12 V batteries. (a) When switch S in Fig. 4a is closed, what is the change in the electric potential V1 across resistor 1, or does V1 remain the same? (b) When switch S in Fig. 4b is closed, what is the change in V1 across resistor 1, or does V1 remain the same?
 |
| Fig.4 |
Answer;
Known:
(a) By the loop rule. It remains the same. This question is aimed at student conceptualization of voltage many students confuse the concepts of voltage and current with the conclusion of this problem.
(b) the loop rule still applies, of coynse, but (by the junction rule and Ohm’s law) the voltage across R1 and R3 (which were the same when the switch was open) are no longer equal. More current is now being supplied by the battery, which means more current is in R3, implying its voltage drop has increased (in magnitude). Thus, by the loop rule (since the battery voltage has not changed) the voltage across R1 was 6.0 V (easily seen from symmetry considerations). With the switch closed, R1 and R2 are equivalent to 3.0 Ω, which means the total load on the battery is 9.0 Ω.
The current therefore is 1.33 A, which implies that the voltage drop across R3 is 8.0 V. The loop rule then tells us that the voltage drop across R1 is
12 V – 8.0 V = 4.0 V.
This is a decrease of 2.0 volts from the value it had when the switch was open.
So ∆V1,final – ∆ V1, initial = 4.0 V – 6.0 V = –2.0 V
(a) By the loop rule. It remains the same. This question is aimed at student conceptualization of voltage many students confuse the concepts of voltage and current with the conclusion of this problem.
(b) the loop rule still applies, of coynse, but (by the junction rule and Ohm’s law) the voltage across R1 and R3 (which were the same when the switch was open) are no longer equal. More current is now being supplied by the battery, which means more current is in R3, implying its voltage drop has increased (in magnitude). Thus, by the loop rule (since the battery voltage has not changed) the voltage across R1 was 6.0 V (easily seen from symmetry considerations). With the switch closed, R1 and R2 are equivalent to 3.0 Ω, which means the total load on the battery is 9.0 Ω.
The current therefore is 1.33 A, which implies that the voltage drop across R3 is 8.0 V. The loop rule then tells us that the voltage drop across R1 is
12 V – 8.0 V = 4.0 V.
This is a decrease of 2.0 volts from the value it had when the switch was open.
So ∆V1,final – ∆ V1, initial = 4.0 V – 6.0 V = –2.0 V
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