JEE Main 2021

LEVELJEE Main

Animated Solution for Physics - Physics and Measurement: Two resistors $R_{1} = (4 \pm 0.8) Ω$ and $R_{2} = (4 \pm 0.4) Ω$ are connected in parallel. The equivalent resistance of their parallel combination will be

Q: What if the resistors were in series?

If they were in series, the formula would be $R_{eq} = R_1 + R_2$. The error would simply be the direct sum of the absolute errors: $\Delta R_{eq} = \Delta R_1 + \Delta R_2$.

Select Answer:

Visualized Solution

$R_{1}$ and $R_{2}$ in Parallel

$R_{1} = (4 \pm 0.8) Ω$
$R_{2} = (4 \pm 0.4) Ω$

Equivalent Resistance Formula

$\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$

Calculating $R_{e q}$

$\frac{1}{R _{e q}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4}$
$R_{e q} = 2 Ω$

Error Propagation Formula

$\frac{Δ R _{e q}}{R _{e q}^{2}} = \frac{Δ R _{1}}{R _{1}^{2}} + \frac{Δ R _{2}}{R _{2}^{2}}$

Substituting Values

$\frac{Δ R _{e q}}{2 ^{2}} = \frac{0.8}{4 ^{2}} + \frac{0.4}{4 ^{2}}$

Calculating $Δ R_{e q}$

$\frac{Δ R _{e q}}{4} = \frac{0.8 + 0.4}{16}$
$Δ R_{e q} = 4 \times \frac{1.2}{16}$
$Δ R_{e q} = 0.3 Ω$

Final Answer

$R_{e q} \pm Δ R_{e q} = (2 \pm 0.3) Ω$

What if they were in series?

$R_{ser i es} = R_{1} + R_{2}$
$Δ R_{ser i es} = Δ R_{1} + Δ R_{2}$

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The Sigma Insight: Errors in Measurement

Solution Diagram

The problem asks us to find the equivalent resistance of two resistors connected in parallel, along with the absolute error in the equivalent resistance. This is a classic application of error analysis in physics.

Analyzing the Setup

We are given two resistors:

R_{1} = (4 \pm 0.8) Ω

R_{2} = (4 \pm 0.4) Ω

These resistors are connected in parallel. Our goal is to find the equivalent resistance

R_{e q}

and its absolute error

Δ R_{e q}

The Master Equation

For two resistors in parallel, the equivalent resistance is given by the reciprocal sum formula:

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

Let's first calculate the main value of the equivalent resistance by substituting the nominal values of

R_{1}

and

R_{2}

\frac{1}{R _{e q}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}

So, the equivalent resistance is:

R_{e q} = 2 Ω

Error Propagation

Now comes the tricky part: finding the error

Δ R_{e q}

. When a formula involves reciprocals, the standard way to find the error is by differentiating the equation.

Differentiating

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

gives:

- \frac{Δ R _{e q}}{R _{e q}^{2}} = - \frac{Δ R _{1}}{R _{1}^{2}} - \frac{Δ R _{2}}{R _{2}^{2}}

Since errors always add up to give the maximum possible error, we drop the negative signs:

\frac{Δ R _{e q}}{R _{e q}^{2}} = \frac{Δ R _{1}}{R _{1}^{2}} + \frac{Δ R _{2}}{R _{2}^{2}}

Final Calculation

Let's substitute the known values into our error equation:

\frac{Δ R _{e q}}{2 ^{2}} = \frac{0.8}{4 ^{2}} + \frac{0.4}{4 ^{2}}

\frac{Δ R _{e q}}{4} = \frac{0.8}{16} + \frac{0.4}{16}

\frac{Δ R _{e q}}{4} = \frac{1.2}{16}

Multiplying both sides by 4, we get:

Δ R_{e q} = 4 \times \frac{1.2}{16} = \frac{1.2}{4} = 0.3 Ω

Conclusion: The equivalent resistance of the parallel combination, along with its error limits, is $(2 \pm 0.3) Ω$ .

Common Traps & FAQs

Why do we differentiate the formula to find the error?

Differentiating a formula is a mathematical trick to find how small changes (errors) in the input variables affect the final result. It directly gives us the relationship between the absolute errors.

Why do we drop the negative signs after differentiation?

In error analysis, we always want to find the *maximum possible error*. Therefore, we assume the worst-case scenario where all errors add up, so we take the absolute values and add them.

Can we use the formula

R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}

to find the error?

Yes, you can, but it's much more complicated. You would have to apply the quotient rule and product rule for errors, which involves relative errors and is highly prone to calculation mistakes. Differentiating the reciprocal formula is much cleaner.

What if the resistors were in series?

If they were in series, the formula would be

R_{e q} = R_{1} + R_{2}

. The error would simply be the direct sum of the absolute errors:

Δ R_{e q} = Δ R_{1} + Δ R_{2}

Why is the equivalent resistance smaller than both individual resistors?

In a parallel circuit, adding more resistors provides more paths for the current to flow, which effectively decreases the overall opposition (resistance) to the current.

Similar Questions

JEE Main 2021

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The resistance $R = \frac{V}{I}$ , where $V = (50 \pm 2) V$ and $I = (20 \pm 0.2) A$ . The percentage error in $R$ is $x$ %. The value of $x$ to the nearest integer is……… .

LEVELBoard

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are $3%$ each, then error in the value of resistance of the wire is

(A)

6%

(B)

zero

(C)

1%

(D)

3%

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Two capacitors with capacitance values $C_{1} = 2000 \pm 10 pF$ and $C_{2} = 3000 \pm 15 pF$ are connected in series. The voltage applied across this combination is $V = 5.00 \pm 0.02 V$ . The percentage error in the calculation of the energy stored in this combination of capacitors is _______.

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In the experiment of Ohm's law, a potential difference of $5.0 V$ is applied across the end of a conductor of length $10.0 cm$ and diameter of $5.00 mm$ . The measured current in the conductor is $2.00 A$ . The maximum permissible percentage error in the resistivity of the conductor is

A physical quantity $z$ depends on four observables $a, b, c$ and $d$ , as $z = \frac{a ^{2} b ^{2/3}}{c d ^{3}}$ . The percentages of error in the measurement of $a, b, c$ and $d$ are $2%$ , $1.5%$ , $4%$ and $2.5%$ respectively. The percentage of error in $z$ is

(A)

13.5%

(B)

16.5%

(C)

14.5%

(D)

12.25%

JEE Main 2013

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The current voltage relation of diode is given by $I = (e^{1000 V / T} - 1)$ mA, where the applied voltage $V$ is in volt and the temperature $T$ is in kelvin. If a student makes an error measuring $\pm 0.01$ V while measuring the current of $5$ mA at $300$ K, what will be the error in the value of current in mA?

A wire has a mass $(0.3 \pm 0.003) g$ , radius $(0.5 \pm 0.005) mm$ and length $(6 \pm 0.06) cm$ . The maximum percentage error in the measurement of its density is

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5%$ and $1%$ , the maximum error in determining the density is

(A)

2.5%

(B)

3.5%

(C)

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(D)

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In order to determine the Young's modulus of a wire of radius $0.2 cm$ (measured using a scale of least count $= 0.001 cm$ ) and length $1 m$ (measured using a scale of least count $= 1 mm$ ), a weight of mass $1 kg$ (measured using a scale of least count $= 1 g$ ) was hanged to get the elongation of $0.5 cm$ (measured using a scale of least count $0.001 cm$ ). What will be the fractional error in the value of Young's modulus determined by this experiment?

(A)

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(B)

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(C)

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(D)

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JEE Advanced 2018

LEVELJEE Advanced

Comprehension Passage

If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation

z = x / y

. If the errors in

x, y

and

z

are

Δ x, Δ y

and

Δ z

respectively, then

z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) (1 \pm \frac{Δ y}{y})^{- 1}

The series expansion for

(1 \pm \frac{Δ y}{y})^{- 1}

to first power in

Δ y / y

, is

1 \mp (Δ y / y)

. The relative errors in independent variables are always added. So, the error in

z

will be

Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y})

The above derivation makes the assumption that

Δ x / x ≪ 1

Δ y / y ≪ 1

. Therefore, the higher powers of these quantities are neglected.

Question 1:

Consider the ratio $r = \frac{( 1 - a )}{( 1 + a )}$ to be determined by measuring a dimensionless quantity $a$ . If the error in the measurement of $a$ is $Δ a$ ( $Δ a / a ≪ 1$ ), then what is the error $Δ r$ in determining $r$ ?

(A)

\frac{Δ a}{( 1 + a ) ^{2}}

(B)

\frac{- 2Δ a}{( 1 + a ) ^{2}}

(C)

\frac{2Δ a}{( 1 - a ) ^{2}}

(D)

\frac{2 a Δ a}{( 1 - a ^{2} )}

Question 2:

In an experiment, the initial number of radioactive nuclei is 3000. It is found that $1000 \pm 40$ nuclei decayed in the first 1.0 s. For $∣ x ∣ ≪ 1$ , $ln (1 + x) = x$ up to first power in $x$ . The error $Δ λ$ , in the determination of the decay constant $λ$ in $s^{- 1}$ , is

Select Answer:

Visualized Solution

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Equivalent Resistance Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Req​

Error Propagation Formula

Substituting Values

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Final Answer

What if they were in series?

The Sigma Insight: Errors in Measurement

Analyzing the Setup

The Master Equation

Error Propagation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The resistance R=IV​, where V=(50±2) V and I=(20±0.2) A. The percentage error in R is x %. The value of x to the nearest integer is……… .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A wire has a mass (0.3±0.003) g, radius (0.5±0.005) mm and length (6±0.06) cm. The maximum percentage error in the measurement of its density is

Comprehension Passage

$R_{1}$ and $R_{2}$ in Parallel

Calculating $R_{e q}$

Calculating $Δ R_{e q}$

The resistance $R = \frac{V}{I}$ , where $V = (50 \pm 2) V$ and $I = (20 \pm 0.2) A$ . The percentage error in $R$ is $x$ %. The value of $x$ to the nearest integer is……… .

A wire has a mass $(0.3 \pm 0.003) g$ , radius $(0.5 \pm 0.005) mm$ and length $(6 \pm 0.06) cm$ . The maximum percentage error in the measurement of its density is