Example 12.9 - Chapter 12 Class 10 - Electricity (Term 2)

Last updated at Aug. 6, 2019 by Teachoo

If in Fig. 12.12, R
_{
1
}
= 10 Ω, R
_{
2
}
= 40 Ω, R
_{
3
}
= 30 Ω, R
_{
4
}
= 20 Ω, R
_{
5
}
= 60 Ω,and a 12 V battery is connected to the arrangement. Calculate

Example 12.9 If in Fig. 12.12, R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω, R5 = 60 Ω, 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.
Given,
R1 = 10 Ω
R2 = 40 Ω
R3 = 30 Ω
R4 = 20 Ω
R5 = 60 Ω
Potential difference = 12 V
Finding total Resistance (R)
Here,
R1 and R2 are connected in parallel
and R3, R4 & R5 are connected in parallel
Suppose we replace R1 and R2 with an equivalent resistor R’
and we replace R3, R4 and R5 with an equivalent resistor R’’
In parallel connection,
1/𝑅_𝑝 = 1/𝑅_1 + 1/𝑅_2 + 1/𝑅_3 + … … …
Calculating R’
1/𝑅′ = 1/𝑅_1 + 1/𝑅_2
1/𝑅′ = 1/10 + 1/40
1/𝑅′ = (4 + 1)/40
1/𝑅′ = 5/40
1/𝑅′ = 1/8
R’ = 8 Ω
Calculating R’’
1/𝑅^′′ = 1/𝑅_3 + 1/𝑅_4 + 1/𝑅_5
1/𝑅^′′ = 1/30 + 1/20 + 1/60
1/𝑅^′′ = (2 + 3 +1)/60
1/𝑅^′′ = 6/60
1/𝑅^′′ = 1/10
R’’ = 10 Ω
Replacing R1 and R2 by R’
and Replacing R3, R4 and R5 by R’’
Now, R’ and R’’ are connected in series.
In series connection,
Rs = R1 + R2 + … …
∴ Net Resistance = R = R’ + R’’
= 8 Ω + 10 Ω
= 18 Ω
(b) Finding total current
By ohm’s law,
V = I R
𝑉/𝑅 = I
I = 𝑉/𝑅
I = 12/18
I = 2/3
I = 0.67 A
Total resistance in circuit is 18 Ω
Total current in circuit is 0.67 A

Article by

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 12 years and a teacher from the past 16 years. He teaches Science, Accounts and English at Teachoo