# Numericals on series and parallel combination of resistors class 10

In this post, we have solved a set of Numericals on series and parallel combinations of resistors for class 10 physics.

Formulas Used

Resistors in series: R_{eq} = R1 + R2 + R3

Resistors in parallel: 1/R_{eq} = 1/R1 + 1/R2 + 1/R3

Numericals on series and parallel combination of resistors for class 10

**Question 1)** If five resistances, each of value 0.5 ohm, are connected in series, what will be the resultant resistance?

**Solution:**As per the law of combination of resistances in series,

R=R1+ R2+ R3+ R4+ R5

R=0.5+0.5+0.5+0.5+0.5=2.5 ohm.

**Question 2**) If 5 resistances of 5 ohm each are connected in parallel, what will be their total resistance?

**Solution**:

1/R = 1/5 +1/5 + 1/5 +1/5 + 1/5 = 5/5 = 1 ohm

=> R = 1/1 ohm = 1 ohm

Equivalent resistance = 1 ohm

**Question 3**) Show how you would connect two 4 ohm resistors to produce a combined resistance of

(a) 2 ohms

(b) 8 ohms.

**Solution :**

(a) By connecting in parallel:

Since equivalent resistance will be

1/ R = 1/4 + 1/4 = 2/4 = 1/2

Therefore, R = 2 ohm

(b) By connecting in series:

Since equivalent resistance will be

R = 4 ohm + 4 ohm = 8 ohm.

**Question 4**) A wire that has resistance R is cut into two equal pieces. The two parts are joined in parallel. What is the resistance of the combination?**Solution :**

The resistance of each part is R/2.

These parts are joined in parallel.

Equivalent resistance R_{eq} is given by

1/R_{eq} = 1/[R/2] + 1/[R/2]

=>1/R_{eq}= 2/R+2/R

R_{eq}=R/4. [answer]

**Question 5**) A battery of 9 V is connected in **series **with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 6 Ω. How much current would flow through the 6 Ω resistor?

**Solution**:

V = 9 V

R_{eq} =.2 +.3 + .4 + .5 + 6 ohm =7.4 ohm

As these resistors are in series, same current will flow through all of these.

current I =V/R_{eq} =9/7.4 A = 1.22 A

**Question 6**) An electric bulb of resistance 30 Ω and a resistance wire of 4 Ω are connected in series with a 6 V battery. Calculate :

(a) total resistance of the circuit.

(b) current through the circuit.

(c) potential difference across the electric bulb.

(d) potential difference across the resistance wire.

**Solution**:

a) Total resistance R = 30 + 4 = 34 ohm

b) V = 6 V

Current through the circuit = I

I = V/R = 6/34 =0.176 A

c) Pd across the electric bulb = I R_{bulb} = 0.176 x 30 V = 5.29 V

d) Pd across the resistance wire =Battery emf – drop across bulb= 6 – 5.29 = 0.71 V

**Question 7**) A 4 Ω coil and a 2 Ω coil are connected in parallel. What is their combined resistance ? A total current of 3 A passes through the coils. What current passes through the 2 Ω coil ?

Solution:

**Question 8**) Two resistances when connected in parallel give the resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

1/R1 + 1/R2 = 1/2

=> 2 = R1R2/[R1 + R2]

=> 2[R1 + R2] = R1R2 ……….. (1)

R1 + R2 = 9 …… (2)

from (1) and (2) we get: 2×9 = R1R2

=> R1R2 = 18

=> R1 = 18/R2 ……..(3)

from (2) & (3)

18/R2 + R2 =9

=>18 + R_{2}^{2} =9R_{2}

=>R_{2}^{2} -9R_{2} + 18 = 0

Solving: R_{2} = 3 or 6

When R_{2} = 3, then from equation (2) we get R_{1} = 6

When R_{2} = 6, then from equation (2) we get R_{1} = 3

So two resistances are 6 ohm and 3 ohm.

**Question 9**) A resistor of 8 ohms is connected in parallel with another resistor X. The resultant resistance of the combination is 4.8 ohms. What is the value of the resistor X?

Solution:

1/8 + 1/X = 1/4.8

=>1/X = 1/4.8 – 1/8 = 3.2/[4.8×8]

=>X = 4.8×8/3.2=12 ohm

**Question 10**) You are given three resistances of 1,2 and 3 ohms. Show by diagrams, how with the help of these resistances you can get:

(i) 6 Ω (ii) 611 Ω (iii) 1.5 Ω

**Question 11**) How will you connect three resistors of 2 Ω, 3 Ω, and 5 Ω respectively so as to obtain a resultant resistance of 2.5 Ω? Draw the diagram to show the arrangement.

**Question **12) How will you connect three resistors of resistances 2 Ω, 3 Ω, and 6Ω obtain a total n si stance of: (a) 4 Ω., and (b) 1Ω?**Solution :**