GATE 2026 Measurements and Instrumentation Practice Questions and Solutions

An analogue voltmeter has a range of 0–150 V and a guaranteed accuracy of \(\pm 1\%\) of full-scale deflection. The pointer settles at 60 V. The limiting (worst-case) error in this particular reading, expressed as a percentage of the reading, is _____ %.

Solution

An accuracy quoted “of full scale” fixes the absolute uncertainty once and for all, independent of where the pointer sits. That absolute band is

Referred to the actual reading of 60 V, the relative limiting error becomes

This is why a deflecting instrument should always be operated in the upper part of its scale: at 150 V the same 1.5 V band would be only 1 % of the reading, but at 15 V it would be 10 %.

A current \(i(t) = 4 + 3\sin(\omega t)\) A flows through a series combination of an ideal permanent-magnet moving-coil (PMMC) ammeter and an ideal moving-iron (MI) ammeter. Both instruments are calibrated to read correctly on their natural quantity. The readings of the PMMC and the MI instruments are, respectively,

1. 4.00 A and 4.53 A
2. 4.53 A and 4.00 A
3. 4.00 A and 5.00 A
4. 5.00 A and 4.53 A

Solution

The deflecting torque of a PMMC movement is proportional to the instantaneous current, and the inertia of the moving system averages that torque. A PMMC therefore indicates the average value:

A moving-iron movement develops a torque proportional to \(i^2\) (the iron pieces magnetise alike whatever the polarity), so it indicates the rms value. For a DC term plus a sinusoid the rms values combine in quadrature:

A Maxwell inductance–capacitance bridge is energised at 1 kHz. Arm 1 holds the unknown coil, modelled as an inductance \(L_1\) in series with a resistance \(R_1\). Arm 2 is a pure resistance \(R_2 = 400\,\Omega\) and arm 3 is a pure resistance \(R_3 = 600\,\Omega\). Arm 4 is a resistance \(R_4 = 12~\text{k}\Omega\) connected in parallel with a capacitance \(C_4 = 0.5\,\mu\text{F}\). Arms 1 and 4 are opposite arms of the bridge, and so are arms 2 and 3. At balance the value of \(L_1\) is _____ mH.

Solution

Balance of a four-arm bridge requires the products of opposite-arm impedances to be equal, \(Z_1 Z_4 = Z_2 Z_3\). Here

Substituting and clearing the denominator of \(Z_4\):

Equating real parts and imaginary parts separately gives the two balance conditions:

Both conditions are free of \(\omega\), which is the practical merit of this bridge: the balance does not drift with supply frequency. Numerically,

As a check on the range of validity, the coil quality factor is \(Q = \omega L_1 / R_1 = \omega C_4 R_4 = 2\pi(1000)(0.5\times10^{-6})(12\times10^{3}) = 37.7\), so \(R_4\) stays a comfortable, realisable value. A Maxwell bridge is suited to medium-\(Q\) coils; for \(Q\) above roughly 10 the required \(R_4\) grows large, and a Hay bridge is preferred instead.

A basic slide-wire DC potentiometer is standardised against a standard cell of emf 1.0186 V by setting the sliding contact at 101.86 cm from the zero end and adjusting the series rheostat until the galvanometer nulls. Keeping the rheostat untouched, the potentiometer is then used to measure the voltage drop across a 0.1 \(\Omega\) four-terminal standard resistor that carries an unknown steady current. A null is obtained at 84.5 cm. The unknown current is _____ A.

Solution

Standardisation fixes the working voltage gradient along the slide wire. With the standard cell balanced at 101.86 cm,

At null the potentiometer draws no current from the source under test, so the measured drop is the true open-circuit value:

The current follows from Ohm's law applied to the standard resistor, whose potential terminals carry no current and so contribute no lead drop:

The secondary winding of a current transformer must never be left open-circuited while its primary carries load current. The reason is that

1. the primary current collapses to zero and the protected line is de-energised
2. the entire primary ampere-turns act as magnetising ampere-turns, driving the core into deep saturation and inducing a dangerously high peak secondary voltage
3. the ratio error becomes negative and the relay under-reaches
4. the secondary burden impedance falls to zero and the winding overheats

Solution

In normal service the secondary ampere-turns very nearly cancel the primary ampere-turns, so the net mmf acting on the core is only the small magnetising component:

Open-circuiting the secondary forces \(I_s = 0\) while \(I_p\) is held by the power circuit, so the whole of \(N_p I_p\) becomes magnetising mmf. The core saturates, the flux waveform flat-tops, and \(d\phi/dt\) becomes very large during the rapid transitions through zero. Since the secondary emf is

and \(N_s\) is large in a step-down current transformer, peak secondary voltages of several kilovolts appear across the open terminals. The insulation and the operator are both at risk, and the heavy core loss can damage the transformer.

A current transformer is rated 500/5 A. With 400 A flowing in the primary, the secondary current is measured as 3.95 A. Taking the nominal ratio as the marked ratio, the percentage ratio error of the transformer at this loading is _____ %.

Solution

The nominal (marked) ratio is

The actual transformation ratio at this loading is

Ratio error is defined as the departure of the nominal ratio from the actual ratio, referred to the actual ratio:

The same figure follows from the measurement viewpoint, comparing the primary current inferred from the marked ratio with the true primary current:

The negative sign says the secondary current is smaller than the marked ratio would predict, which is the expected direction: the magnetising and core-loss components of the exciting current are supplied at the expense of the secondary ampere-turns.

The vertical amplifier of an oscilloscope has a Gaussian frequency response with a 3 dB bandwidth of 20 MHz, for which the rise time obeys \(t_r = 0.35/\text{BW}\). A pulse whose true 10–90 % rise time is 12 ns is applied to the input. The rise time displayed on the screen is _____ ns.

Solution

First find the rise time the instrument itself would show for an ideal step input:

For cascaded stages with Gaussian (non-overshooting) responses, the rise times add in the root-sum-square sense, because the equivalent Gaussian widths add in quadrature under convolution:

The displayed edge is markedly slower than the real one. A useful rule follows from the same relation: to keep the measurement error below about 2 %, the instrument rise time should be no more than one-fifth of the signal rise time, which here would call for a bandwidth of at least 146 MHz.

A digital storage oscilloscope operating in real-time sampling mode has its sampling rate set to 5 kSa/s, and its anti-alias filter is switched off. A pure sinusoid of 6 kHz is applied to the vertical input. The frequency of the waveform reconstructed on the screen is

1. 1 kHz
2. 5 kHz
3. 6 kHz
4. 11 kHz

Solution

Sampling at \(f_s\) replicates the input spectrum about every integer multiple of \(f_s\). A component at \(f\) therefore appears in the baseband at

Here the Nyquist limit is \(f_s/2 = 2.5\) kHz, and the 6 kHz input lies well above it, so aliasing is unavoidable. With \(k = 1\),

Any other \(k\) gives a larger value (\(k=0\) gives 6 kHz, \(k=2\) gives 4 kHz), so the screen shows a perfectly convincing 1 kHz sinusoid that does not exist at the probe tip. This is exactly why the sampling rate on a DSO must be checked against the fastest component expected, not against the component of interest.

A dual-slope integrating digital voltmeter is clocked at 100 kHz. Its run-up (signal integration) interval is fixed at 1000 clock pulses, and the reference voltage used for run-down is 5 V. For a steady unknown DC input the run-down interval is counted as 640 clock pulses. The unknown input voltage is _____ V.

Solution

During run-up the integrator output ramps for the fixed interval \(T_1 = N_1/f_{clk}\) at a rate set by the input. Starting from zero, the integrator output at the end of run-up is

The reference of opposite polarity is then switched in and the integrator ramps back to zero in the counted interval \(T_2 = N_2/f_{clk}\):

Equating the two expressions, \(R\), \(C\) and the clock period all cancel, which is the whole point of the dual-slope scheme — component drift and clock drift do not affect the reading:

The 100 kHz clock fixes only the absolute timing: the run-up lasts \(1000/10^{5} = 10\) ms, which is one full period of a 100 Hz supply ripple and therefore also gives good series-mode rejection of 50 Hz hum.

A Pt100 resistance thermometer follows the linear law \(R_T = R_0\left(1 + \alpha T\right)\) with \(R_0 = 100\,\Omega\) at 0 °C and \(\alpha = 0.00385\;/^\circ\text{C}\). It is wired to the readout in a two-wire configuration, each of the two connecting leads having a resistance of 0.5 \(\Omega\). The readout measures the total resistance at its terminals and converts it to temperature using the same linear law with the same \(R_0\) and \(\alpha\). When the sensor is at 150 °C, the error in the indicated temperature is _____ °C.

Solution

The true sensor resistance at 150 °C is

In a two-wire connection both leads lie in series with the element, so the readout sees

Inverting the linear law gives the indicated temperature:

Because the law is linear, the error can be written directly as \(\Delta T = R_{lead}/(R_0\alpha) = 1.0/0.385 = 2.60\) °C, a fixed offset that is independent of the temperature being measured. Eliminating it is the reason industrial RTDs are wired in three-wire or four-wire form.

For a compact revision of these instrument models see the Measurements and Instrumentation revision notes.