In an electrical circuit two resistances $R_{1} = (4 \pm 0.5)$ and $R_{2} = (12 \pm 0.5)$ are connected in parallel. What will be the equivalent resistance with percentage error?
( Given the effective resistance $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ )

3 \pm 22.92

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3 \pm 18.6

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16 \pm 22.92

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16 \pm 18.6

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Solution

The correct option is A $3 \pm 22.92$ %
The effective resistance, $\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$R = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$

$R = \frac{4 \times 12}{16} = 3$ ohm
For error, $l n R = l n R_{1} + l n R_{2} - l n (R_{1} + R_{2})$

Differentiating: $\frac{Δ R}{R} = \frac{Δ R_{1}}{R_{1}} + \frac{Δ R_{2}}{R_{2}} + \frac{Δ (R_{1} + R_{2})}{R_{1} + R_{2}}$

Maximum % error $= \frac{Δ R}{R} \times 100$

$= (\frac{0.5}{4} + \frac{0.5}{12} + \frac{1}{16}) 100$
$= 12.5 + 4.17 + 6.25$
$= 22.92$

$R = 3 \pm 22.92$ %

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Similar questions

r_{1} = (5.0 \pm 0.2) Ω

r_{2} = (10.0 \pm 0.1) Ω

R_{1} (4 \pm

Ω

R_{2} (12 \pm

Ω

R = R_{1} + R_{2}

R_{1} = (3 \pm 0.09) Ω

R_{2} = (6 \pm 0.36) Ω

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