In a circuit potential difference across resistor V=(4±0.25)V and current in resistor, I=(1±0.1)A. The value of resistance with percentage error is

R=Vi=41=4Ω ΔRR×100=(ΔVV+ΔII)100 =(0.254+0.11)100 =6.25+10=16.25%

In a circuit potential difference across resistor V=(4±0.25)V and current in resistor, I=(1±0.1)A. The value of resistance with percentage error is

R=Vi=41=4Ω ΔRR×100=(ΔVV+ΔII)100 =(0.254+0.11)100 =6.25+10=16.25%

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In a circuit potential difference across resistor $V = (4 \pm 0.25) V$ and current in resistor, $I = (1 \pm 0.1) A$ . The value of resistance with percentage error is

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$4 Ω \pm 22.25 %$

$4 Ω + 16.25 %$

$4 Ω + 18.25 %$

$(4 \pm 0.4) Ω$

answer is B.

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$R = \frac{V}{i} = \frac{4}{1} = 4 Ω$

$\frac{Δ R}{R} \times 100 = (\frac{Δ V}{V} + \frac{Δ I}{I}) 100$

$= (\frac{0.25}{4} + \frac{0.1}{1}) 100$

$\begin{array}{l} = 6.25 + 10 \\ = 16.25 % \end{array}$

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In a circuit potential difference across resistor V=(4±0.25)V and current in resistor, I=(1±0.1)A. The value of resistance with percentage error is

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In a circuit potential difference across resistor $V = (4 \pm 0.25) V$ and current in resistor, $I = (1 \pm 0.1) A$ . The value of resistance with percentage error is