Question

7. Find the effective resistance of the circuit when three resistors of 6
, 12
and 24
are

connected in (i) series, (ii) parallel

Answer 1

Explanation:

in series

R=R1+R2+R3

R=6+12+24=

R=42 ohm

in parallel

1/R=1/R1+1/R2+1/R3

1/R = 1/6+1/12+1/24

1/R=42

Answer 2

$\large\bf\underline {To \: find:-}$

• Resistance of given resistors when connected in
• (i) series
• (ii) parellel

$\huge\bf\underline{Solution:-}$

$\bf\underline{\red{Given:-}}$

Resistance of three resistors :-

• R1 = 6Ω
• R2 = 12Ω
• R3 = 24Ω

Effective resistance :-

When Resistors are connected in series

→ Rs = R1 + R2 + R3

→ Rs = 6Ω + 12Ω + 24Ω

→ Rs = 42Ω

When Resistors are connected in parallel

→ 1/Rp = 1/R1 + 1/R2 + 1/R3

→ 1/Rp = 1/6 + 1/12 + 1/24

→ 1/Rp = (4 + 2 + 1)/24

→ 1/Rp = 7/24

→ 7Rp = 24

→ Rp = 24/7

→ Rp = 3.42Ω

Hence,

• Equivalent resistance when Resistors are connected in series is 42Ω

• Equivalent resistance when Resistors are connected in parallel is 3.42Ω

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