Resistors in Series and Parallel MCQ

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Resistors in series and parallel are the two ways of connection in a circuit. A resistor is an electrical component that provides electrical resistance which can be used to leverage the amount of current flowing through the circuit. This topic has more weightage in Class 12 Chapter 3 Current Electricity. In a series connection, the resistors are connected end to end whereas, in a parallel connection, the resistors are connected in parallel to one another.

The total resistance of resistors in series is given by,

R_total = R₁ + R₂ + ….. + R_n

The total resistance of resistors in parallel is given by,

\(\frac{1}{R_{Total}}= \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3} +...\frac{1}{R_n}\)

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Resistors in Series and Parallel MCQ

Ques 1. In series connection of resistors, what happens to the current across each resistor?

a) Increases

b) Decreases

c) Remain the same

d) Initially increases and then decreases

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Answer: c

Explanation: When the resistors are connected in series, and current is passed through them, the current passing through each of the resistors is the same. This is because the resistors are connected end to the end and, therefore, there is only one path for the current to flow through.

Ques 2. Identify the combination which is not a series connection.

a) Resistance box

b) Decorative bulbs

c) Fuses

d) Domestic appliances

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Answer: d

Explanation: Domestic appliances in a house are connected in parallel combinations, and not in series combinations. This arrangement is done so that each of the appliances can switch on and off independently and does not alter the connection with the other appliances.

Ques 3. The equivalent overall resistance in a parallel connection is smaller than the series connection.

a) True

b) False

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Answer. a

Explanation: As we all know the total resistance of resistors in a series circuit is Rtotal = R1 + R2 + ….. + Rn, whereas for a parallel circuit, it is, \(\frac{1}{R_{Total}}= \frac{1}{R_1}+ \frac{1}{R_2} + \frac{1}{R_3} +...\frac{1}{R_n}\). Since in parallel circuit, it is the reciprocal of the resistances, it tends to be smaller than series connection.

Ques 4. Calculate resultant resistance between the points A and B in the circuit shown in the figure below.

Calculate resultant resistance between the points A and B in the circuit

a) 4 ohms

b) 2 ohms

c) 8 ohms

d) 6 ohms

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Answer: b

Explanation: We can simplify the given diagram as follows:

From the diagram above, we can take the middle resistors connected in series and then calculate all the other resistors connected in parallel.

\(\frac{1}{R_1} = \frac{1}{4} + \frac{1}{8} = \frac{4+6}{4 * 6} = \frac {10}{24}\) → R₁ = \(\frac{24}{10}\) = 2.4

\(\frac{1}{R_2} = \frac{1}{2} + \frac{1}{8} = \frac{2+8}{2 * 8} = \frac {10}{16}\)→ R₂ = \(\frac{16}{10}\) = 1.6

Now, R1 and R2 are in series, so, Rs = 2.4 + 1.6 = 4.0.

8 ohms, Rs, and the bottom 8 ohms are in parallel, so, 1/Rp = ⅛ + ¼ + ⅛ = ½

On taking the inverse of Rp, we get, Rp = 2 ohms.

Ques 5. Two wires of the same material have the same length but their radii are in the ratio of 5:3. They are combined in series, where the resistance of the thicker wire is 12 ohms. Calculate the total resistance of the combination.

a) 40

b) 12

c) 32

d) 20

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Answer: c

Explanation: The given ratio of radii = 5:3;

R2/R1 = 5/3

R2 = 5/3 x R1

R1 = 12 ohms (given)

R2 = 5/3 x 12

R2 = 20

Total resistance of resistors connected in series Rs = R1 + R2

Rs = 20 + 12

Rs = 32 ohms

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Ques 6. Two resistors are connected in parallel, whose resistance values are in the ratio 3:1. Find the ratio of power dissipated.

a) 1:3

b) 3:1

c) 1:2

d) 2:1

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Answer: a

Explanation: We know that,

Power = voltage²/ resistance

Since the resistors are connected in parallel, the voltage across them will be the same.

From this relation, power and resistance are inversely proportional to each other.

Thus, P1/P2 = R2/R1 =1/3

So, the power dissipated is in the ratio of 1:3.

Ques 7. A circuit has three similar resistors, each of 20 ohms resistance. Two of them are connected in parallel, and this combination is connected in series with the third one. The maximum power that can be consumed by each resistor is 30 W. Then, what is the maximum power that can be consumed by the combination of all three resistors?

a) 30

b) 20

c) 35

d) 45

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Answer: d

Explanation: The equivalent overall resistance of the parallel combination is:

1/R1 = 1/20 + 1/20 = 2/20 = 1/10

On taking reciprocal,

R1 = 10 ohms.

R1 is in series with R2; So, R3 = R1 + R2 = 10 + 20 = 30 ohms.

Now, we can employ the method of cross-multiplication:

For 20 ohms resistor → 30 W power consumed

For 30 ohms resistor combination → x

20x = 30 × 30

x = 30×30/ 20

x = 45

Therefore, the power consumed by the parallel combination is 45 ohms.

Ques 8. Which connection is preferable to connect bulbs?

a) Series

b) Parallel

c) Both series and parallel

d) Neither series nor parallel

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Answer: b

Explanation: Bulbs are connected in parallel so that even if one of the bulbs blows out, the others continue to get a current supply. This is why parallel connection is used in domestic electrical circuits.

Ques 9. Batteries are generally connected?

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

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Answer: a

Explanation: Batteries are generally connected in series so that we can obtain the desired voltage since voltages add up once they are connected in series.

Ques 10. In a _________ circuit, the total resistance is greater than the largest resistance in the circuit.

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

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Answer: a

Explanation: In series circuits, the total resistance is the sum of all the resistance in the circuit, hence the total is greater than the largest resistance.

Ques 11. In a ____________ circuit, the total resistance is smaller than the smallest resistance in the circuit.

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

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Answer: b

Explanation: Since the equivalent resistance is 1/sum of the reciprocals of all the resistances in the circuit, it is smaller than the smallest resistance in the circuit.

Ques 12. Which connection is the most cost-efficient connection?

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

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Answer: a

Explanation: The advantage of series connection is that they share the supply voltage, hence cheap low voltage appliances may be used.

Ques 13. Calculate the equivalent resistance between A and B.

Calculate the equivalent resistance between A and B.

a) 2 ohm

b) 4 ohm

c) 6 ohm

d) 8 ohm

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Answer: a

Explanation: The 2 and the 3 ohm resistors are in series.

R1 = 3 + 2 = 5

The equivalent of these two resistors is in parallel with the 5 ohm resistor.

1/R2 = 1/R1 + ⅕

= ⅕ + ⅕

= ⅖

R2 = 5/2

The equivalent of these three resistances is in series with the 1.5 ohm resistor.

R3 = R2 + 1.5

R3 = 5/2 + 1.5

R3 = 2.5 + 1.5

R3 = 4 ohm

Finally, the equivalent of these resistances is in parallel with the 4 ohm resistor.

1/R4 = ¼ + ¼

1/R4 = 2/4 = ½

R4 = 2 ohm

Ques 14. Calculate the equivalent resistance between A and B.

Calculate the equivalent resistance between A and B

a) 6.67 ohm

b) 46.67 ohm

c) 26.67 ohm

d) 10.67 ohm

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Answer: b

Explanation: The three 20 ohm resistors in the middle are in parallel to each other.

1/R1 = 1/20 + 1/20 + 1/20 = 3/20

R1 = 20/3

The rest of the resistors are in series.

R2 = 20 + 20 + 20/3

R2 = 46.67 ohm

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Resistors in Series and Parallel MCQ

Resistors in Series and Parallel MCQ

CBSE CLASS XII Related Questions

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

6.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

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Resistors in Series and Parallel MCQ

Resistors in Series and Parallel MCQ

CBSE CLASS XII Related Questions

1.Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

3.A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).

6.Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Three batteries E1, E2, and E3 of emfs and internal resistances (4 V, 2 \(\Omega\)), (2 V, 4 \(\Omega\)) and (6 V, 2 \(\Omega\)) respectively are connected as shown in the figure. Find the values of the currents passing through batteries E1, E2, and E3.

2.
A current carrying circular loop of area A produces a magnetic field \( B \) at its centre. Show that the magnetic moment of the loop is \( \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \).

6.
Two point charges \( q_1 = 16 \, \mu C \) and \( q_2 = 1 \, \mu C \) are placed at points \( \vec{r}_1 = (3 \, \text{m}) \hat{i}\) and \( \vec{r}_2 = (4 \, \text{m}) \hat{j} \). Find the net electric field \( \vec{E} \) at point \( \vec{r} = (3 \, \text{m}) \hat{i} + (4 \, \text{m}) \hat{j} \).