GATE 2026 EE question paper PDF and answer key

The core flux is set by $V/f$: India $\frac{220}{50} = 4.4$, USA $\frac{110}{60} = 1.83$. So the peak flux density in the USA is about 42% of the Indian value.

• (A) True: lower flux gives a smaller magnetizing (no-load) current.
• (B) False: core losses fall and copper loss is unchanged, so losses are lower, not higher.
• (C) False: the peak flux density is lower.
• (D) False.

The two AND gates give $\bar{x}y$ and $\bar{x}z$, so the OR output is $\bar{x}(y + z)$. The NOR gives $\bar{x}\bar{y}\bar{z}$. $\text{OUT} = \bar{x}\bar{y}\bar{z} \cdot \bar{x}(y + z) = \bar{x}\,\bar{y}\bar{z}(y + z) = 0$ The output is constant 0.

The C transistor pulls node N (10 kΩ load) low, so $N = \bar{C}$. OUT is pulled low when $A$ and $B$ are both on, or when N is high: $\text{OUT} = \overline{AB + \bar{C}} = \overline{AB}\,C$ Maximum power: OUT low draws $\frac{5}{5\text{k}} = 1$ mA and N low draws $\frac{5}{10\text{k}} = 0.5$ mA, so $P = 5 \times 1.5\ \text{mA} = 7.5$ mW.

For $R_L = R_{Th}$ at port 2, short the 10 V source. That also shorts the 5 Ω, leaving $10\ \Omega \parallel 4\ \Omega$: $R_{Th} = \frac{10 \times 4}{14} \approx 2.86\ \Omega$

Diode conducts for $\pi/6<\theta<5\pi/6$; $I_{avg}=(1/2\pi)\int(10\sin\theta-5)d\theta=1.09$ A. Therefore, the answer is 1.09 A (official accepted range 1.04 to 1.14).