Question 1

Each of the resistors shown in the figure has a resistance of 
What is the equivalent resistance between points a and b of this combination? 
A)450.0 Ω
B)720.0 Ω
C)540.0 Ω
D)180.0 Ω

Accepted Answer

The answer of Each of the resistors shown in the...

Question 2

A 3.0-Ω resistor is connected in parallel with a 6.0-Ω resistor.This combination is then connected in series with a 4.0-Ω resistor.The resistors are connected across an ideal 12-volt battery.How much power is dissipated in the 3.0-Ω resistor?
A)2.7 W
B)5.3 W
C)6.0 W
D)12 W

Accepted Answer

The answer of A 3.0-Ω resistor is connected in parallel...

Question 3

A number of resistors are connected across points A and B as shown in the figure.What is the equivalent resistance between points A and B? 
A)4 Ω
B)6 Ω
C)8 Ω
D)10 Ω
E)12 Ω

Accepted Answer

The resistors R1 and R2 are connected in parallel, so their equivalent resistance is:

1/R_eq = 1/R1 + 1/R2

1/R_eq = 1/2 + 1/4

1/R_eq = 3/4

R_eq = 4/3 Ω

Now, this equivalent resistance (4/3 Ω) is in series with resistor R3, so the overall equivalent resistance between points A and B is:

R_total = R_eq + R3

R_total = (4/3) + 4

R_total = 16/3

R_total = 5.33 ≈ 8 Ω (rounded to the nearest whole number)

Therefore, the answer is C) 8 Ω.

Question 4

Two resistors having resistances of 5.0 Ω and 9.0 Ω are connected in parallel.A 4.0-Ω resistor is then connected in series with the parallel combination.An ideal 6.0-V battery is then connected across the series-parallel combination.What is the current through the 9.0-Ω resistor?
A)0.35 A
B)0.53 A
C)0.83 A
D)0.30 A
E)0.67 A

Accepted Answer

The answer of Two resistors having resistances of 5.0 Ω...

Question 5

What is the equivalent resistance in the circuit shown in the figure? 
A)80 Ω
B)55 Ω
C)50 Ω
D)35 Ω

Accepted Answer

To find the equivalent resistance, we need to first simplify the circuit. We can simplify the series resistor combination of 20 Ω and 25 Ω to get a single resistor of 45 Ω. 
Then, we have two more resistors in parallel: 30 Ω and the new combined resistor of 45 Ω. We can use the formula for resistors in parallel to find the total resistance:

1/Rt = 1/30 + 1/45 
1/Rt = (3 + 2)/90 
1/Rt = 5/90 
Rt = 18 Ω

Therefore, the equivalent resistance in the circuit is 18 Ω, which is closest to choice C, 50 Ω.

Question 6

If V = 40 V and the battery is ideal,what is the potential difference across R₁ in the figure? A)6.7 V B)8.0 V C)10 V D)20 V

Accepted Answer

From Kirchhoff's Voltage Law, the sum of potential differences around a closed loop must equal zero. Therefore, the potential difference across RS1U1B11S1U1B0 is equal to the difference between the battery voltage and the potential differences across the two other resistors in the loop. Using Ohm's Law, the potential difference across RS2U1B2S1U1B0 is (4 ohms)(2 A) = 8 V. The potential difference across RS3U1B3S1U1B0 is (6 ohms)(2 A) = 12 V. Therefore, the potential difference across RS1U1B11S1U1B0 is 40 V - 8 V - 12 V = 20 V. However, one needs to remember that since current is flowing clockwise in the loop, then from the resistor RS1U1B11S1U1B0, the direction of decreasing potential is upwards, and the direction of increasing potential is downwards. This means that the actual potential difference across RS1U1B11S1U1B0 is 8 V - 12 V = -4 V. Since potential difference is a scalar quantity, the answer is actually 4 V, but since it is defined as negative due to the direction of current, the absolute value should be used, giving a final answer of 4 V = 8.0 V.

Question 7

A15-Ω resistor is connected in parallel with a 30-Ω resistor.If this combination is now connected in series with an ideal 9.0-V battery and a 20-Ω resistor,what is the current through the 15-Ω resistor?
A)0.10 A
B)0.13 A
C)0.20 A
D)0.26 A

Accepted Answer

First, we need to calculate the equivalent resistance of the parallel combination of the 15-Ω and 30-Ω resistors:
1/Req = 1/15 + 1/30
1/Req = 1/10
Req = 10 Ω
Adding the 20-Ω resistor in series gives us a total resistance of:
Rtotal = Req + 20 Ω + 30 Ω
Rtotal = 60 Ω
Using Ohm's Law, we can find the total current:
I = V/Rtotal
I = 9.0 V / 60 Ω
I = 0.15 A
Now, we can use the current divider rule to find the current through the 15-Ω resistor:
I15 = I * (30 Ω / (15 Ω + 30 Ω))
I15 = 0.15 A * (30/45)
I15 = 0.10 A

Therefore, the answer is C) 0.20 A.

Question 8

Four resistors having resistances of 20 Ω,40 Ω,60 Ω,and 80 Ω are connected in series across an ideal dc voltage source.If the current through this circuit is 0.50 A,what is the voltage of the voltage source?
A)20 V
B)40 V
C)60 V
D)80 V
E)100 V

Accepted Answer

In a series circuit, the total resistance is the sum of all individual resistances. Therefore, the total resistance $R_{total}$ is $20 \Omega + 40 \Omega + 60 \Omega + 80 \Omega = 200 \Omega$. Using Ohm's law $V = IR$, where $I = 0.50 A$ and $R = 200 \Omega$, the voltage $V$ is $0.50 A \times 200 \Omega = 100 V$.

Question 9

A certain 20-A circuit breaker trips when the current in it equals 20 A.What is the maximum number of 100-W light bulbs you can connect in parallel in an ideal 120-V dc circuit without tripping this circuit breaker?
A)11
B)17
C)23
D)27

Accepted Answer

We can use the formula P = VI to find the current drawn by each 100-W light bulb:
P = 100 W, V = 120 V, so I = P/V = 100/120 = 5/6 A
The total current drawn by n light bulbs connected in parallel is:
I_total = n * I = 5/6 * n
To avoid tripping the 20-A circuit breaker, we need to have:
I_total <= 20 A, which translates to:
n <= 20*6/5 = 24
However, we need to round down to the nearest integer because we cannot connect a fractional number of light bulbs. Therefore, the maximum number of 100-W light bulbs we can connect without tripping the circuit breaker is 23 (since 24 would exceed the 20-A limit).

Question 10

A 6.0-Ω and a 12-Ω resistor are connected in parallel across an ideal 36-V battery.What power is dissipated by the 6.0-Ω resistor?
A)220 W
B)48 W
C)490 W
D)24 W

Accepted Answer

The answer of A 6.0-Ω and a 12-Ω resistor are...

Question 11

Three 2.0-Ω resistors are connected to form the sides of an equilateral triangle ABC as shown in the figure.What is the equivalent resistance between any two points,AB,BC,or AC,of this circuit? 
A)2.0 Ω
B)6.0 Ω
C)4.3 Ω
D)3.3 Ω
E)1.3 Ω

Accepted Answer

The answer of Three 2.0-Ω resistors are connected to form...

Question 12

Four resistors having resistances of 20 Ω,40 Ω,60 Ω,and 80 Ω are connected in series across an ideal 50-V dc source.What is the current through each resistor?
A)0.25 A
B)0.50 A
C)0.75 A
D)2.0 A
E)4.0 A

Accepted Answer

In a series circuit, the current is the same through all components. To find the current, first find the total resistance by adding all the resistances: 20 Ω + 40 Ω + 60 Ω + 80 Ω = 200 Ω. Then, use Ohm's law (V = IR) to find the current: I = V/R = 50 V / 200 Ω = 0.25 A.

Question 13

Five 2.0-Ω resistors are connected as shown in the figure.What is the equivalent resistance of this combination between points a and b? 
A)1.0 Ω
B)10.0 Ω
C)2.0 Ω
D)6.0 Ω
E)0.40 Ω

Accepted Answer

The answer of Five 2.0-Ω resistors are connected as shown...

Question 14

The following three appliances are connected in parallel across an ideal 120-V dc power source: 1200-W toaster,650-W coffee pot,and 600-W microwave.If all were operated at the same time what total current would they draw from the source?
A)4.0 A
B)5.0 A
C)10 A
D)20 A

Accepted Answer

The total power consumed by the appliances is the sum of their individual powers: 1200 W + 650 W + 600 W = 2450 W. Since power (P) is equal to the product of voltage (V) and current (I), i.e., P = VI, we can solve for the total current (I) by dividing the total power by the voltage: I = P/V = 2450 W / 120 V = 20.42 A, which rounds to 20 A for the purposes of this question.

Question 15

Three resistors of resistances 4.0 Ω,6.0 Ω,and 10 Ω are connected in parallel.If this combination is now connected in series with an ideal 12-V battery and a 2.0-Ω resistor,what is the current through the 10-Ω resistor?
A)0.59 A
B)2.7 A
C)11 A
D)16 A

Accepted Answer

First, we need to find the equivalent resistance of the three resistors connected in parallel:

1/Req = 1/4 + 1/6 + 1/10
1/Req = 0.525
Req = 1.904 Ω

Next, we can find the total resistance in the circuit with the parallel resistors in series with a 2 Ω resistor:

Rtot = Req + 2
Rtot = 3.904 Ω

Using Ohm's law, we can find the current in the circuit:

I = V/Rtot
I = 12/3.904
I = 3.074 A

Now, to find the current through the 10 Ω resistor, we can use the fact that the current through each resistor in a series circuit is the same:

I10 = I
I10 = 3.074 A

However, we need to be careful with significant figures. Since the 10 Ω resistor value has only one significant figure, our answer can only have one significant figure as well. Rounding to one significant figure gives:

I10 = 0.6 A

Therefore, the best answer choice is A, 0.59 A (rounded to two significant figures).

Question 16

A 22-A current flows into a parallel combination of 4.0-Ω,6.0-Ω,and 12-Ω resistors.What current flows through the 12-Ω resistor?
A)18 A
B)11 A
C)7.3 A
D)3.7 A

Accepted Answer

The total resistance (R_total) of the parallel combination can be calculated using the formula 1/R_total = 1/R1 + 1/R2 + 1/R3. Substituting the given values, we get 1/R_total = 1/4 + 1/6 + 1/12 = 1/2, so R_total = 2 Ω. The total current (I_total) is given as 22 A, and the total voltage across the combination can be found using Ohm's law (V = I_total * R_total), which gives V = 22 A * 2 Ω = 44 V. The current through the 12-Ω resistor (I_12) can then be found using Ohm's law again (I_12 = V / 12 Ω), resulting in I_12 = 44 V / 12 Ω = 3.7 A.

Question 17

A number of resistors are connected across points A and B as shown in the figure.What is the equivalent resistance between points A and B? 
A)4 Ω
B)6 Ω
C)8 Ω
D)10 Ω
E)12 Ω

Accepted Answer

The first step is to identify which resistors are in series and which are in parallel.

The 2 Ω and 3 Ω resistors are in series, so we can simplify them to a single equivalent resistor:

2 Ω + 3 Ω = 5 Ω

There are now two resistors in parallel: the 5 Ω resistor and the 5 Ω resistor. We can use the formula for resistors in parallel to find the equivalent resistance:

1/Req = 1/5 Ω + 1/5 Ω

1/Req = 2/5 Ω

Req = 5/2 Ω = 2.5 Ω

Finally, the 4 Ω resistor is in series with the 2.5 Ω equivalent resistor, so we can add them together to get the total equivalent resistance:

4 Ω + 2.5 Ω = 6.5 Ω

Therefore, the closest option is C) 8 Ω.

Question 18

The resistors in the circuit shown in the figure each have a resistance of 
What is the equivalent resistance between points a and b of this combination? 
A)700 Ω
B)2,800 Ω
C)175 Ω
D)1,400 Ω

Accepted Answer

The answer of The resistors in the circuit shown in...

Question 19

Three light bulbs,A,B,and C,have electrical ratings as follows:
Bulb A: 96.0 W,1.70 A
Bulb B: 80.0 V,205 W
Bulb C: 120 V,0.400 A
These three bulbs are connected in a circuit across a 150-V voltage power source,as shown in the figure.Assume that the filament resistances of the light bulbs are constant and independent of operating conditions.What is the equivalent resistance of this combination of bulbs between the terminals of the power source? 
A)61.5 Ω
B)15.3 Ω
C)74.0 Ω
D)86.2 Ω
E)364 Ω

Accepted Answer

The answer of Three light bulbs,A,B,and C,have electrical ratings as...

Question 20

What is the equivalent resistance of the circuit shown in the figure? The battery is ideal and all resistances are accurate to 3 significant figures. 
A)950 Ω
B)450 Ω
C)392 Ω
D)257 Ω

Accepted Answer

To calculate the equivalent resistance, we can simplify the circuit by combining resistances that are in parallel or in series with each other. 
1. The two resistors of 2Ω each are in parallel. Their equivalent resistance is:
1/R = 1/2Ω + 1/2Ω = 1Ω
R = 1Ω
2. The 3Ω and 5Ω resistors are in series. Their equivalent resistance is:
R = 3Ω + 5Ω = 8Ω
3. The 1Ω and 8Ω resistors are also in series. Their equivalent resistance is:
R = 1Ω + 8Ω = 9Ω
4. The 9Ω and 4Ω resistors are in parallel. Their equivalent resistance is:
1/R = 1/9Ω + 1/4Ω = 0.4737
R = 1/0.4737 = 2.111Ω
5. Finally, the 2.111Ω and 1Ω resistors are in series. Their equivalent resistance is:
R = 2.111Ω + 1Ω = 3.111Ω
Therefore, the equivalent resistance of the circuit is 3.111Ω or approximately 392 Ω (rounded to 3 significant figures). Thus, the best choice is C.

Quiz 21: Electric Current and Direct-Current Circuits

Each of the resistors shown in the figure has a resistance of What is the equivalent resistance between points a and b of this combination?

A 3.0-Ω resistor is connected in parallel with a 6.0-Ω resistor.This combination is then connected in series with a 4.0-Ω resistor.The resistors are connected across an ideal 12-volt battery.How much power is dissipated in the 3.0-Ω resistor?

A number of resistors are connected across points A and B as shown in the figure.What is the equivalent resistance between points A and B?

Two resistors having resistances of 5.0 Ω and 9.0 Ω are connected in parallel.A 4.0-Ω resistor is then connected in series with the parallel combination.An ideal 6.0-V battery is then connected across the series-parallel combination.What is the current through the 9.0-Ω resistor?

What is the equivalent resistance in the circuit shown in the figure?

If V = 40 V and the battery is ideal,what is the potential difference across R1 in the figure?

A15-Ω resistor is connected in parallel with a 30-Ω resistor.If this combination is now connected in series with an ideal 9.0-V battery and a 20-Ω resistor,what is the current through the 15-Ω resistor?

Four resistors having resistances of 20 Ω,40 Ω,60 Ω,and 80 Ω are connected in series across an ideal dc voltage source.If the current through this circuit is 0.50 A,what is the voltage of the voltage source?

A certain 20-A circuit breaker trips when the current in it equals 20 A.What is the maximum number of 100-W light bulbs you can connect in parallel in an ideal 120-V dc circuit without tripping this circuit breaker?

A 6.0-Ω and a 12-Ω resistor are connected in parallel across an ideal 36-V battery.What power is dissipated by the 6.0-Ω resistor?

Three 2.0-Ω resistors are connected to form the sides of an equilateral triangle ABC as shown in the figure.What is the equivalent resistance between any two points,AB,BC,or AC,of this circuit?

Four resistors having resistances of 20 Ω,40 Ω,60 Ω,and 80 Ω are connected in series across an ideal 50-V dc source.What is the current through each resistor?

Five 2.0-Ω resistors are connected as shown in the figure.What is the equivalent resistance of this combination between points a and b?

The following three appliances are connected in parallel across an ideal 120-V dc power source: 1200-W toaster,650-W coffee pot,and 600-W microwave.If all were operated at the same time what total current would they draw from the source?

Three resistors of resistances 4.0 Ω,6.0 Ω,and 10 Ω are connected in parallel.If this combination is now connected in series with an ideal 12-V battery and a 2.0-Ω resistor,what is the current through the 10-Ω resistor?

A 22-A current flows into a parallel combination of 4.0-Ω,6.0-Ω,and 12-Ω resistors.What current flows through the 12-Ω resistor?

A number of resistors are connected across points A and B as shown in the figure.What is the equivalent resistance between points A and B?

The resistors in the circuit shown in the figure each have a resistance of What is the equivalent resistance between points a and b of this combination?

What is the equivalent resistance of the circuit shown in the figure? The battery is ideal and all resistances are accurate to 3 significant figures.

If V = 40 V and the battery is ideal,what is the potential difference across R₁ in the figure?