GATE Biomedical Engineering Electrical Circuits Syllabus

How do inductances and capacitances combine in series?

Inductors in series add directly: $L_{eq} = L_{1}+L_{2}+\cdots$. Capacitors in series add reciprocally: $\frac{1}{C_{eq}} = \frac{1}{C_{1}}+\frac{1}{C_{2}}+\cdots$.

State Kirchhoff's Current Law (KCL).

The algebraic sum of currents entering (or leaving) any node is zero: $\sum i = 0$. It is based on conservation of charge — the sum of currents entering a node equals the sum leaving it.

State Kirchhoff's Voltage Law (KVL).

The algebraic sum of all voltages around any closed loop is zero: $\sum v = 0$. It is based on conservation of energy.

What is the voltage-divider formula for two series resistors $R_{1}$ and $R_{2}$ across a source $V$?

$$V_{1} = V\cdot\frac{R_{1}}{R_{1}+R_{2}}$$ The voltage across each resistor is proportional to its resistance.

What is the current-divider formula for two parallel resistors $R_{1}$ and $R_{2}$ carrying total current $I$?

$$I_{1} = I\cdot\frac{R_{2}}{R_{1}+R_{2}}$$ The current splits inversely with resistance.

State the Superposition Theorem.

In a linear circuit with multiple independent sources, the response (voltage or current) in any element equals the algebraic sum of the responses caused by each independent source acting alone, with all other independent sources deactivated.

When applying superposition, how do you 'turn off' a voltage source and a current source?

An independent voltage source is replaced by a short circuit (set to $0\,\text{V}$), and an independent current source is replaced by an open circuit (set to $0\,\text{A}$). Dependent sources are left unchanged.

Why can superposition not be used directly to calculate power?

Because power is a nonlinear function of voltage/current ($P = I^{2}R$ or $V^{2}/R$). Superposition applies only to linear quantities like voltage and current, not to power.

State Thevenin's Theorem.

Any linear two-terminal network can be replaced by an equivalent circuit consisting of a single voltage source $V_{Th}$ in series with a single resistance $R_{Th}$, where $V_{Th}$ is the open-circuit voltage and $R_{Th}$ is the equivalent resistance seen from the terminals.

How do you find $V_{Th}$ and $R_{Th}$ in Thevenin's theorem?

$V_{Th}$ is the open-circuit voltage measured across the load terminals (load removed). $R_{Th}$ is found by deactivating all independent sources and computing the equivalent resistance at the terminals (or as $R_{Th}=V_{oc}/I_{sc}$).

State Norton's Theorem.

Any linear two-terminal network can be replaced by an equivalent circuit consisting of a single current source $I_{N}$ in parallel with a single resistance $R_{N}$, where $I_{N}$ is the short-circuit current and $R_{N}=R_{Th}$.

What is the relationship between Thevenin and Norton equivalents?

They are related by source transformation: $R_{N}=R_{Th}$ and $V_{Th}=I_{N}R_{N}$, equivalently $I_{N}=\frac{V_{Th}}{R_{Th}}$.

How is $R_{Th}$ (or $R_{N}$) found when the network contains dependent sources?

Independent sources are deactivated but dependent sources are kept. A test source ($1\,\text{V}$ or $1\,\text{A}$) is applied at the terminals, and $R_{Th}=\frac{V_{test}}{I_{test}}$; alternatively use $R_{Th}=\frac{V_{oc}}{I_{sc}}$.

State the Maximum Power Transfer Theorem for a DC resistive network.

Maximum power is transferred to the load when the load resistance equals the Thevenin resistance of the source network: $R_{L}=R_{Th}$.

What is the maximum power delivered to the load, and the efficiency at that condition?

$$P_{max}=\frac{V_{Th}^{2}}{4R_{Th}}$$ At maximum power transfer the efficiency is only $50\%$, since equal power is dissipated in $R_{Th}$ and $R_{L}$.

For an AC source with Thevenin impedance $Z_{Th}=R_{Th}+jX_{Th}$, what load gives maximum power transfer?

Maximum power transfer occurs when the load impedance is the complex conjugate of the source impedance: $Z_{L}=Z_{Th}^{*}=R_{Th}-jX_{Th}$.

State the Reciprocity Theorem.

In a linear, bilateral, single-source network, the ratio of the response (current) to the excitation (voltage) remains the same when the positions of the source and the response measurement are interchanged.

What are the conditions (limitations) for applying the Reciprocity Theorem?

The network must be linear and bilateral, contain a single source, and have no dependent sources. Initial conditions are zero. The ratio of excitation to response (transfer resistance) is unchanged on interchanging source and response points.

Define the peak value (amplitude) of an alternating quantity.

The peak value $V_{m}$ (or $I_{m}$) is the maximum instantaneous value the waveform reaches during a cycle, i.e., the amplitude measured from the zero axis to the crest.

Define the average value of a periodic waveform and give it for one half-cycle of a sine wave.

The average value is the mean of the instantaneous values over a period: $V_{avg}=\frac{1}{T}\int_{0}^{T}v(t)\,dt$. For a half-cycle of a sine wave, $V_{avg}=\frac{2V_{m}}{\pi}\approx 0.637\,V_{m}$.

Define the RMS (root mean square) value of an AC waveform.

$$V_{rms}=\sqrt{\frac{1}{T}\int_{0}^{T}v^{2}(t)\,dt}$$ It is the equivalent DC value that produces the same heating (power dissipation) in a resistor.

What is the RMS value of a sinusoidal waveform of peak $V_{m}$?

$$V_{rms}=\frac{V_{m}}{\sqrt{2}}\approx 0.707\,V_{m}$$

Define the form factor and crest factor, and give their values for a sine wave.

Form factor $=\frac{V_{rms}}{V_{avg}}$; for a sine wave $=\frac{0.707}{0.637}\approx 1.11$. Crest (peak) factor $=\frac{V_{m}}{V_{rms}}$; for a sine wave $=\sqrt{2}\approx 1.414$.

Define apparent power, including its formula and unit.

Apparent power is the product of RMS voltage and RMS current: $S=V_{rms}I_{rms}$, measured in volt-amperes (VA). It is the magnitude of complex power.

Define active (real) power, including its formula and unit.

Active power is the actual power consumed: $P=V_{rms}I_{rms}\cos\phi$, measured in watts (W), where $\phi$ is the phase angle between voltage and current. It represents energy dissipated/converted to useful work.

Define reactive power, including its formula and unit.

Reactive power is the power that oscillates between source and reactive elements: $Q=V_{rms}I_{rms}\sin\phi$, measured in volt-amperes reactive (VAR). It does no net work but is needed to sustain magnetic/electric fields.

Write the power triangle relationship among $S$, $P$, and $Q$.

$$S=\sqrt{P^{2}+Q^{2}}$$ with complex power $\vec{S}=P+jQ$. The three quantities form a right triangle where $S$ is the hypotenuse.

Define power factor and express it in terms of $P$ and $S$.

Power factor is the cosine of the angle between voltage and current: $\text{pf}=\cos\phi=\frac{P}{S}$. It ranges from $0$ to $1$, and is lagging for inductive loads and leading for capacitive loads.

How does power factor differ for purely resistive, purely inductive, and purely capacitive loads?

Purely resistive: $\phi=0^{\circ}$, $\text{pf}=1$ (unity), all power is active. Purely inductive: $\phi=+90^{\circ}$, $\text{pf}=0$ lagging, only reactive power. Purely capacitive: $\phi=-90^{\circ}$, $\text{pf}=0$ leading, only reactive power.