GATE 2025 IN Q.25: solution and answer

GATE 2025 · INQ.25 · 1 mark · MCQ

Three parallel admittances $Y_{a} = - j 0.2 S$ , $Y_{b} = 0.3 S$ , and $Y_{c} = j 0.4 S$ connected in parallel with a voltage source $V_{s} = 10∠4 5^{\circ} V$ , draw a total current $I_{s}$ from the source. The currents flowing through each of these admittances are $I_{a}$ , $I_{b}$ , and $I_{c}$ , respectively. Let $I = I_{b} + I_{c}$ . The phase relation between $I$ and $I_{s}$ is

(A)
$I$ leads $I_{s}$ by $19.4 4^{\circ}$
(B)
$I$ lags $I_{s}$ by $19.4 4^{\circ}$
(C)
$I$ leads $I_{s}$ by $33.6 9^{\circ}$
(D)
$I$ lags $I_{s}$ by $33.6 9^{\circ}$

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Answer and solution

ANSWER

$I = 5∠98.1 3^{\circ} A$ and $I_{s} = 3.606∠78.6 9^{\circ} A$ , so $I$ leads $I_{s}$ by $19.4 4^{\circ}$ . Option (A).

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

All three admittances are in parallel across the same source, so each branch current is the source voltage times that branch's admittance: $I_{a} = Y_{a} V_{s}$ , $I_{b} = Y_{b} V_{s}$ , $I_{c} = Y_{c} V_{s}$ (Ohm's law in admittance form, $I = Y V$ ).
Total current (admittances in parallel add): $I_{s} = V_{s} (Y_{a} + Y_{b} + Y_{c}) = V_{s} (- j 0.2 + 0.3 + j 0.4) = V_{s} (0.3 + j 0.2)$ .
Partial current: $I = I_{b} + I_{c} = V_{s} (Y_{b} + Y_{c}) = V_{s} (0.3 + j 0.4)$ .
Both currents carry the same factor $V_{s}$ , so their phase difference is just the difference of the two admittance angles: $∠ (0.3 + j 0.4) = tan^{- 1} \frac{0.4}{0.3} = 53.1 3^{\circ}$ and $∠ (0.3 + j 0.2) = tan^{- 1} \frac{0.2}{0.3} = 33.6 9^{\circ}$ .
The same result with the source angle included ( $∣0.3 + j 0.4∣ = 0.5 S$ , $∣0.3 + j 0.2∣ = 0.13 = 0.3606 S$ ): $I I_{s} = 10∠4 5^{\circ} \times 0.5∠53.1 3^{\circ} = 5∠98.1 3^{\circ} A = 10∠4 5^{\circ} \times 0.3606∠33.6 9^{\circ} = 3.606∠78.6 9^{\circ} A$
Phase difference: $98.1 3^{\circ} - 78.6 9^{\circ} = 19.4 4^{\circ}$ . $I$ has the larger angle, so $I$ leads $I_{s}$ by $19.4 4^{\circ}$ , which is option (A).

KEY IDEA

Every branch sees the same voltage $V_{s}$ , so every current is $V_{s}$ times an admittance. The angle between any two of these currents is therefore the angle between the two admittances, $∠ (Y_{b} + Y_{c}) - ∠ (Y_{a} + Y_{b} + Y_{c})$ . The $4 5^{\circ}$ of the source cancels out.

WATCH OUT FOR

Picking $33.6 9^{\circ}$ (options C or D). That is the angle of $I_{s}$ measured from $V_{s}$ , not from $I$ . Other slips are leaving out $Y_{a} = - j 0.2 S$ when forming $I_{s}$ , and reading the sign the wrong way round: $I$ is at the larger angle, so it leads $I_{s}$ .

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GATE 2025 IN Q.25: solution and answer

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