Concepts

Class 10
Chapter 12 Class 10 - Electricity

Here Different Resistors are connected between 2 points parallel to each other

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Example

This is done to decrease net Resistance of Circuit

In this case Total Resistance of circuit is equal to sum of reciprocals of Individual Resistance of the Resistors

1/R = 1/R 1 + 1/R 2 + 1/R 3

IMPORTANT POINTS

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## When 2 Resistors are Connected in Parallel

1. Different current flow through each resistor but overall Current flowing throughout the circuit remains the Same
Hence we can say that
Total Current = Current flowing through first Resistor + Current flowing through Second Resistor + Current flowing through Third Resistor
I = I 1 + I 2 + I 3

2. Potential Difference across allย  both Resistors remain the same (as they are both between Same points)
Hence we can say that
Potential Diff=V
It is same for both Resistor 1 and Resistor 2

Now we know that as per Ohm's Law

Potential Diff=Current*Resistance

Potential Difference/Resistance=Current

Current=Potential Diff/Resistance

I = V/R

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## How is the Resistance formula Derived - for parallel circuits?

Note - In case of parallel, the total resistance is always less than the resistance of the individual resistors.

The net resistance produced is lowest when resistors are connected in parallel.

## Questions

NCERT Question 5 - How is a voltmeter connected in the circuit to measure the potential difference between two points?

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Q1 Page 216 - Judge the equivalent resistance when the following are connected in parallel โ
(a) 1 โฆ and 10 6 โฆ,
(b) 1 โฆ and 10 3 โฆ, and 10 6 โฆ.

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Q2 Page 216 - An electric lamp of 100 โฆ, a toaster of resistance 50 โฆ, and a water filter of resistance 500 โฆ are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

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Q4 Page 216 - How can three resistors of resistances 2 โฆ, 3 โฆ, and 6 โฆ be connected to give a total resistance of (a) 4 โฆ, (b) 1 โฆ?

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Q5 Page 216 - What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 โฆ, 8 โฆ, 12 โฆ, 24 โฆ?

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Example 12.8 - In the circuit diagram given in Fig. 12.10, suppose the resistors R 1 , R 2 and R 3 have the values 5 โฆ, 10 โฆ, 30 โฆ, respectively, which have been connected to a battery of 12 V. Calculate
(a) the current through each resistor,
(b) the total current in the circuit, and
(c) the total circuit resistance.

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Example 12.9 - If in Fig. 12.12, R 1 = 10 โฆ, R 2 = 40 โฆ, R 3 = 30 โฆ, R 4 = 20 โฆ, R 5 = 60 โฆ, and a 12 V battery is connected to the arrangement. Calculate
(a) the total resistance in the circuit, and
(b) the total current flowing in the circuit.

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Question 5 - What is the minimum resistance which can be made using five resistors each of 1/5 ฮฉ?

(a) 1/5 ฮฉ

(b) 1/25 ฮฉ

(c) 1/10 ฮฉ

(d) 25 ฮฉ

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### Transcript

Example If 2 Resistors - 3 ฮฉ and 4 ฮฉ are connected in parallel. What is the resistance of the circuitSince resistors are connected in parallel, Resistance of Circuit is given by 1/๐ = 1/๐_1 + 1/๐_2 1/๐ = 1/3 + 1/4 1/๐ = (4 + 3)/(3 ร 4) 1/๐ = 7/12 R = 12/7 R = 1.71 ฮฉ How is the Resistance formula Derived - for Parallel circuits? In parallel circuit, Voltage is same Let Voltage in circuit = V Now, Total Current = Current in R1 + Current in R2 + Current in R3 Current in R1 V = I1 ร R1 ๐/๐_1 = I1 I1 = ๐/๐_1 Current in R2 V = I2 ร R2 ๐/๐_2 = I2 I2 = ๐/๐_2 Current in R3 V = I3 ร R3 ๐/๐_3 = I3 I3 = ๐/๐_3 Also, Total Voltage = Total Current ร Total Resistance Since Voltage in Circuit = Total Voltage = V V = Total Current ร Total Resistance ๐/(๐๐๐ก๐๐ ๐๐๐ ๐๐ ๐ก๐๐๐๐) = Total Current Total Current = ๐/(๐๐๐ก๐๐ ๐๐๐ ๐๐ ๐ก๐๐๐๐) Now, Total Current = I1 + I2 + I3 ๐/(๐๐๐ก๐๐ ๐๐๐ ๐๐ ๐ก๐๐๐๐) = ๐/๐_1 + ๐/๐_2 + ๐/๐_3 ๐/(๐๐๐ก๐๐ ๐๐๐ ๐๐ ๐ก๐๐๐๐) = V(1/๐_1 + 1/๐_2 + 1/๐_3 ) ๐/(๐ป๐๐๐๐ ๐น๐๐๐๐๐๐๐๐๐) = ๐/๐น_๐ + ๐/๐น_๐ + ๐/๐น_๐