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

(a) the total resistance in the circuit, and

(b) the total current flowing in the circuit.

Slide28.JPG

Slide29.JPG
Slide30.JPG
Slide31.JPG
Slide32.JPG Slide33.JPG

Go Ad-free

Transcript

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

Maninder Singh's photo - Co-founder, Teachoo

Made by

Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 14 years and a teacher from the past 18 years. He teaches Science, Economics, Accounting and English at Teachoo