Two resistances are expressed as R 1 4- 0-5

The correct option is A (16± 0.5 ...

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Standard XII

Physics

Power in an AC Circuit

Two resistanc...

Question

Two resistances are expressed as $R_{1} (4 \pm$ 0.5 %) $Ω$ and $R_{2} (12 \pm$ 0.5 %) $Ω$ . The net resistance when they are connected in series with percentage error is:
(In series $R = R_{1} + R_{2}$ )

(16 \pm

0.5 %)

Ω

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(16 \pm

6.25 %)

Ω

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(16 \pm

22 %)

Ω

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(16 \pm

2.2 %)

Ω

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Solution

The correct option is A $(16 \pm$ 0.5 %) $Ω$
$R = R_{1} + R_{2} = 4 + 12 = 16 Ω$
$R_{1} = 4 \pm (0.5$ % of $4)$ $Ω$
$R_{2} = 12 \pm (0.5$ % of $12)$ $Ω$
Errors will be added
$R = (R_{1} + R_{2}) + Δ R_{1} + Δ R_{2}$
$= 16 \pm (0.5$ % of $4) + (0.5$ % of $12)$
$= 16 \pm (0.5$ % of $16)$
$R = 16 \pm 0.5$ % $Ω$

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Q. If two resistors of resistances R1 = (4±0.5) ohm and R2 = (16±0.5) ohm are connected in series, find the equivalent resistance with limits of percentage error.

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Two resistances are expressed as R1(4± 0.5 %)Ω and R2(12± 0.5 %)Ω. The net resistance when they are connected in series with percentage error is:(In series R=R1+R2)

The correct option is A (16± 0.5 %) ΩR=R1+R2=4+12=16ΩR1=4±(0.5 % of 4) ΩR2=12±(0.5 % of 12) ΩErrors will be addedR=(R1+R2)+ΔR1+ΔR2 =16±(0.5 % of 4)+(0.5 % of 12) =16±(0.5 % of 16)R=16±0.5 % Ω

Two resistances are expressed as $R_{1} (4 \pm$ 0.5 %) $Ω$ and $R_{2} (12 \pm$ 0.5 %) $Ω$ . The net resistance when they are connected in series with percentage error is:
(In series $R = R_{1} + R_{2}$ )