Electronics (ECE) - MCQ Practice Questions

Understanding:

We need to determine the number of machine cycles for the 8085 instruction LDA 2050H.

Step 1: Identify the instruction format

LDA is a 3-byte instruction: one opcode byte followed by two address bytes (low byte then high byte of the 16-bit address).

Step 2: Count machine cycles

The execution of LDA 2050H involves the following machine cycles:

1. Opcode Fetch (OF): Fetch the opcode byte from memory.

2. Memory Read (MR): Read the low-order byte of the address (50H).

3. Memory Read (MR): Read the high-order byte of the address (20H).

4. Memory Read (MR): Read the data from the address 2050H into the Accumulator.

Total machine cycles = $4$

Step 3: Count T-states for verification

LDA uses $13$ T-states: OF = 4, MR = 3, MR = 3, MR = 3 $\Rightarrow$ 4 + 3 + 3 + 3 = 13 T-states, consistent with 4 machine cycles.

Answer:

The LDA 2050H instruction requires 4 machine cycles.

Quick Tip:

For LDA/STA/LHLD/SHLD, always count 4 machine cycles: 1 opcode fetch + 2 address byte reads + 1 data operation. This is a commonly tested fact.

Understanding:

We need to identify which flag controls auto-increment or auto-decrement of SI and DI during 8086 string operations.

Step 1: 8086 String Instructions

String instructions such as MOVSB, MOVSW, LODSB, STOSB, CMPSB, and SCASB use SI (Source Index) and DI (Destination Index) registers, which are automatically updated after each operation.

Step 2: The Direction Flag (DF)

Step 3: Eliminate other options

Answer:

The Direction Flag (DF) controls whether SI and DI are incremented or decremented during string operations.

Quick Tip:

Remember CLD (Clear Direction Flag) → auto-increment (forward processing) and STD (Set Direction Flag) → auto-decrement (backward processing). This distinction is frequently tested.

Understanding:

We need to find how many $8\text{KB}$ blocks fit in the $64\text{KB}$ address space of the 8085, given that the upper 3 address bits select each block.

Formula:

Step 1: Compute using total address space and block size

Step 2: Verify using the address bit argument

The 8085 has a 16-bit address bus. With the upper 3 bits (A15, A14, A13) used for chip select:

Answer:

The $64\text{KB}$ space is divided into $8$ blocks of $8\text{KB}$ each.

Quick Tip:

$n$ upper address bits always yield $2^n$ blocks. With 3 bits: $2^3=8$ blocks. Each block size = Total space $/2^n$.

Understanding:

We need to find the magnetic field intensity $H$ corresponding to an electric field $E_0=10\text{V/m}$ for a plane wave in free space.

Formula:

For a uniform plane wave in free space, the ratio of electric to magnetic field intensity equals the intrinsic impedance:

Step 1: Recall the intrinsic impedance of free space

Step 2: Solve for $H$

Answer:

The magnetic field intensity is approximately $0.0265\text{A/m}$.

Quick Tip:

${\eta }_0=120\pi \approx 377\Omega$ is a standard value to memorise. A common mistake is inverting the ratio and computing $E/H=H$ instead of $H=E/{\eta }_0$.

Understanding:

We must find the electric field magnitude due to a point charge at a given distance.

Formula:

Coulomb's law gives the electric field due to a point charge:

Step 1: Substitute values

Answer:

The magnitude of the electric field at $r=2\text{m}$ is $9\text{V/m}$.

Quick Tip:

$\frac{1}{4\pi {\epsilon }_0}=9\times 10^9$ is a standard constant. The field falls as $1/r^2$, so doubling the distance reduces the field to one-quarter.

Understanding:

We need the magnetic flux density outside an infinitely long current-carrying conductor.

Formula:

Using Ampere's circuital law, the magnetic flux density at distance $r$ from an infinite straight wire is:

Step 1: Substitute values

Answer:

The magnetic flux density at $r=5\text{cm}$ is $40\mu \text{T}$.

Quick Tip:

Note that $\pi$ cancels cleanly in this formula, making the arithmetic simple. A common error is forgetting to convert centimetres to metres before substituting.

Understanding:

We must find the capacitance of a parallel-plate capacitor with a dielectric.

Formula:

The capacitance of a parallel-plate capacitor with a dielectric is:

Step 1: Substitute values

Answer:

The capacitance of the parallel-plate capacitor is approximately $354\text{pF}$.

Quick Tip:

The free-space capacitance (without dielectric) would be $\approx 88.5\text{pF}$; inserting a dielectric of ${\epsilon }_r=4$ multiplies this by exactly $4$.

Understanding:

We need to apply the differential (point) form of Gauss's law to find $\nabla \cdot \mathbf{D}$.

Formula:

The point form of Gauss's law in electrostatics is:

Step 1: Direct application

Step 2: Note the distinction

The relation $\mathbf{D}={\epsilon }_0\mathbf{E}$ in free space gives $\nabla \cdot \mathbf{E}={\rho }_v/{\epsilon }_0$, but the fundamental Maxwell equation involves $\mathbf{D}$, not $\mathbf{E}$, so no ${\epsilon }_0$ appears here.

Answer:

The divergence of $\mathbf{D}$ equals the volume charge density directly.

Quick Tip:

A very common trap is confusing $\nabla \cdot \mathbf{D}={\rho }_v$ with $\nabla \cdot \mathbf{E}={\rho }_v/{\epsilon }_0$. The ${\epsilon }_0$ factor belongs with $\mathbf{E}$, not with $\mathbf{D}$.

Understanding:

We must identify the correct electromagnetic boundary condition for the tangential component of $\mathbf{E}$ at a dielectric-dielectric interface with no free surface charges or currents.

Formula:

From Faraday's law applied at the interface (tangential $\mathbf{E}$ comes from $\oint \mathbf{E}\cdot d\mathbf{l}=0$ for static fields):

Step 1: Derive from integral form

Applying $\oint \mathbf{E}\cdot d\mathbf{l}=0$ around a small rectangular contour straddling the interface and letting the height shrink to zero gives:

Step 2: Distinguish from the normal component condition

The normal component of $\mathbf{D}$ (not $\mathbf{E}$) satisfies the boundary condition:

The relation ${\epsilon }_1{E}_{t1}={\epsilon }_2{E}_{t2}$ applies to normal $D$ components and is a common distractor.

Answer:

The tangential electric field is continuous across any dielectric boundary.

Quick Tip:

Remember: tangential $\mathbf{E}$ is always continuous; normal $\mathbf{D}$ is continuous when there is no free surface charge. The permittivities ${\epsilon }_1,{\epsilon }_2$ appear only in the normal-component condition.

Understanding:

We need the wavelength of a plane wave propagating in free space at a given frequency.

Formula:

The wavelength is related to frequency and wave speed by:

Step 1: Substitute values

Answer:

The wavelength at $3\text{GHz}$ in free space is $0.1\text{m}$, i.e., $10\text{cm}$.

Quick Tip:

A quick rule: at $3\text{GHz}$, $\lambda =10\text{cm}$. Doubling frequency halves the wavelength. This range ($1$–$30\text{GHz}$) is the microwave band.

Understanding:

We must find the intrinsic impedance of a lossless dielectric medium.

Formula:

The intrinsic impedance of a lossless medium is:

Step 1: Substitute values

Answer:

The intrinsic impedance of the medium is approximately $125.7\Omega$.

Quick Tip:

With ${\mu }_r=1$, higher ${\epsilon }_r$ always lowers $\eta$ below ${\eta }_0$. The speed of light in this medium is also reduced by $\sqrt{{\epsilon }_r{\mu }_r}=3$, giving $v=c/3$.

Understanding:

We need to find the $x$-component of $\mathbf{E}$ from the given scalar potential $V$.

Formula:

The electric field is the negative gradient of the potential:

Step 1: Compute partial derivatives

Step 2: Evaluate $E_x$ at $(1,2,3)$

Answer:

The $x$-component of the electric field at $(1,2,3)$ is $-6\text{V/m}$.

Quick Tip:

The negative sign in $\mathbf{E}=-\nabla V$ is frequently dropped by mistake, giving $+6\text{V/m}$. Always remember that $\mathbf{E}$ points in the direction of decreasing potential.