Resistors in Series and Parallel Important Questions

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Resistors in Series and Parallel Important Questions

Ques 1. How are batteries connected? (2 marks)

Ans. Batteries are connected in series. A battery is a series of cells that are connected in series. 

Ques 2. Why is the resistance of a parallel circuit less than the resistance of individual resistors? (2 marks)

Ans. The resistance of a parallel circuit is less than the resistance of individual resistors because the total resistance of resistors connected in parallel is equal to the reciprocal of the sum of the total resistance. 

Ques 3. Calculate the equivalent resistance for the circuit which is connected to 24 V battery and also find the potential difference across 4 Ω and 6 Ω resistors in the circuit. (3 marks)

Ans. Since the resistors are connected in series, the total resistance in the circuit is given by,

4 Ω + 6 Ω = 10 Ω

Current I in the circuit= V/ Req = 24/10 = 2 .4 A

Voltage across 4Ω resistor

V1 = IR1 = 2 . 4 × 4 = 9.6V

Voltage across 6 Ω resistor

V2 = IR1 = 2 . 4× 6 =14 .4V

Ques 4. Calculate the equivalent resistance in the following circuit and also find the current I, I1 and I2 in the given circuit. (3 marks)

Ans. The resistors are connected in parallel. Therefore the total resistance is given by,

1/R = 1/R1 + 1/R2

1/R = ¼ + ⅙ 

1/R = 5/12

R = 12/5 ohms

The resistors are connected in parallel, and the potential (voltage) across each resistor is the same.

I₁ = \(\frac{V}{R_1} = \frac{24 V}{6 \Omega} = 6A\)

I₂ = \(\frac{V}{R_2} = \frac{24}{6} = 4A\)

The current I is the total of the currents in the two branches. Then,

I = I1 + I2= 6 A + 4 A = 10 A

Ques 5. When two resistances connected in series and parallel their equivalent resistances are 15 Ω and 56/15 Ω respectively. Find the individual resistances. (5 marks)

Ans. Rs = R1 + R2 = 15 Ω…(1)

\(R_P = \frac{R_1R_2}{R_1 + R_2} = \frac{56}{15} \Omega\)...(2)

From equation (1) substituting R1 + R2 in equation (2)

\( \frac{R_1R_2}{15} = \frac{56}{15} \Omega\)

R1R2 = 56

R2 = 56/15…(3)

Substituting for R2 in equation (1) from equation (3)

R1 + 56/ R1 = 15

Then,

\(\frac{R^2_1 + 56}{R_1} = 15\)

R²₁ + 56 = 15R₁

R₁ - 15R₁ + 56 = 0

The above equation can be solved using factorisation.

R12-8 R1-7 R1+ 56 = 0

R1 (R1– 8) – 7 (R1– 8) = 0

(R1– 8) (R1– 7) = 0

If (R1= 8 Ω)

using in equation (1)

8 + R2 = 15

R2 = 15 – 8 = 7 Ω ,

R2 = 7 Ω i.e , (when R1 = 8 Ω ; R2 = 7 Ω)

If (R1= 7 Ω)

Substituting in equation (1)

7 + R2 = 15

R2 = 8 Ω , i.e , (when R1 = 8 Ω ; R2 = 7 Ω)

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Ques 6. Calculate the equivalent resistance between A and B in the given circuit. (5 marks)

Ans. First, we will calculate the parallel resistance of the three units and then calculate the resistances in series.

1st Unit:

1/R1 = ½ + ½ 

1/R1 = 1 ohm

R1 = 1 ohm

2nd Unit:

1/R2 = ¼ + ¼ 

1/R2 = 2/4 = ½ ohm

R2 = 2 ohm

3rd Unit:

1/R3 = ⅙ + ⅙ 

1/R3 = 2/6 = ⅓

R3 = 3 ohm

Total Resistance in series Rs = R1 + R2 + R3

Rs = 1 + 2+ 3

Rs = 6 ohms

Ques 7. Five resistors are connected in the configuration as shown in the figure. Calculate the equivalent resistance between the points a and b. (5 marks)

Ans. Since all the resistances in the outside loop are the same (1Ω), the current in the branches ac and ad must be equal. So the electric potential at point c and d is the same hence no current flows into 5 Ω resistance. It implies that the 5 Ω has no role in determining the equivalent resistance and it can be removed. So the circuit is simplified as shown in the figure.

Unit 1:

Since the resistors are in parallel,

1/R1 = 1/1 + 1/1

1/R1 = 2

R1 =½

Unit 2:

Since the resistors are in parallel,

1/R2 = 1/1 + 1/1

1/R2 = 2

R2 =½

Total Resistance in series Rs = R1 + R2 

Rs = ½ + ½ 

Rs = 1 ohm

Ques 8. A circuit has a resistor with a resistance of 3Ω followed by three parallel branches, each holding a resistor with a resistance of 5Ω. What is the total equivalent resistance of the circuit? (3 marks)

Ans. First, we need to calculate the equivalent resistance of the three resistors in parallel. To do this, we will use the following equation:

\(\frac{1}{R_{eq}}\)= ∑ \(\frac{1}{R} = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{3}{5}\)

R_eq = 5/3 ohms

Now, to get the total equivalent resistance, we can simply add the two remaining values, since they are in series:

Rs = 3 + 5/3 =14/3 ohms

Ques 9. A voltage is applied across points A and B so that current flows from A, to R2, to B. What is the value of this voltage if the current through R2 is 4A? (5 marks)

R1=3Ω

R2=2Ω

Ans. First, we need to calculate the current flow through R2 without the extra voltage attached. We will need to calculate the total equivalent resistance of the circuit. Since the two resistors are in series, we can simply add them.

R_eq =R1+R2=3Ω+2Ω=5Ω

Then, we can use Ohm's law to calculate the current through the circuit:

V=IR

I=V/R = 12/5 =2.4A

Now that we have the current, we can calculate the additional current that the new voltage contributes:

I_tot=I+I_new

I_new=4A

I_new=4A−2.4A=1.6A

There is only one resistor (R2) in the path of the new voltage, so we can calculate what that voltage needs to be to deliver the new current:

V=IR=(1.6A)(2Ω)=3.2V

Ques 10. Two lightbulbs, one graded at 40W and one graded at 60W are connected in series to a battery. Which one will be brighter? What if they are connected in parallel? (3 marks)

Ans. First we have to find out how the resistances of light bulb correlate to the power rating. For a resistor, the power dissipated is:

P = IV = V²/R

Thus, there is an inverse relationship between the resistance of the lightbulb and the power rating. In a series connection, they share the same current. Whereas in a parallel they share the same voltage. Thus, for the two lightbulbs in series, the one with the higher resistance (lower wattage) will be brighter, and for a parallel configuration the one with the lower resistance (higher wattage) will be brighter.

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