GATE 2022 EE Q.18: solution and answer

GATE 2022 · EEQ.18 · 1 mark · MCQ

A balanced Wheatstone bridge $A B C D$ has the following arm resistances:

$R_{A B} = 1 k Ω \pm 2.1%$ ; $R_{B C} = 100 Ω \pm 0.5%$ ; $R_{C D}$ is an unknown resistance; $R_{D A} = 300 Ω \pm 0.4%$ . The value of $R_{C D}$ and its accuracy is

(A)
$30 Ω \pm 3 Ω$
(B)
$30 Ω \pm 0.9 Ω$
(C)
$3000 Ω \pm 90 Ω$
(D)
$3000 Ω \pm 3 Ω$

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Answer and solution

ANSWER

$R_{C D} = 30 Ω \pm 0.9 Ω$ , option (B).

Official answer: BSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

The arms go round the loop $A \to B \to C \to D \to A$ , so arm $A B$ is opposite arm $C D$ and arm $B C$ is opposite arm $D A$ .
Balance condition. With the supply across $A$ – $C$ and the detector across $B$ – $D$ , no detector current flows when $B$ and $D$ are at the same potential, which means the two dividers split the supply in the same ratio: $\frac{R _{B C}}{R _{A B} + R _{B C}} = \frac{R _{C D}}{R _{D A} + R _{C D}} \Rightarrow R_{A B} R_{C D} = R_{B C} R_{D A}$ In words, the products of opposite arms are equal. Putting the supply across $B$ – $D$ instead gives the same condition.
$R_{C D} = \frac{R _{B C} R _{D A}}{R _{A B}} = \frac{100 \times 300}{1000} Ω = 30 Ω$
Limiting (worst-case) error of a product or quotient: the relative (percentage) errors of all the factors add, whether a factor multiplies or divides. For $y = \frac{ab}{c}$ , $\frac{δ y}{y} = \pm (\frac{δ a}{a} + \frac{δ b}{b} + \frac{δ c}{c})$ .
$\frac{δ R _{C D}}{R _{C D}} = \pm (0.5 + 0.4 + 2.1) % = \pm 3%$
Absolute error: $δ R_{C D} = 3% \times 30 Ω = 0.03 \times 30 Ω = \pm 0.9 Ω$ .
So $R_{C D} = 30 Ω \pm 0.9 Ω$ , which is option (B).

KEY IDEA

At balance, the products of opposite arms of a Wheatstone bridge are equal: $R_{A B} R_{C D} = R_{B C} R_{D A}$ . When the unknown is a product or quotient of the other arms, its limiting percentage error is the sum of their percentage errors. You then turn that percentage into ohms using the computed value.

WATCH OUT FOR

Pairing the wrong arms. Writing $R_{C D} = R_{A B} R_{D A} / R_{B C} = 3000 Ω$ gives options (C) or (D). Another trap is reading the $\pm 3%$ relative error as $\pm 3 Ω$ , which gives option (A). A third is subtracting the error of the divisor arm, $2.1 - 0.5 - 0.4 = 1.2%$ : in a limiting-error calculation, relative errors always add.

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GATE 2022 EE Q.18: solution and answer

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