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Moving Coil Galvanometer & Conversions — MH-CET Physics MCQs with Solutions

Free MH-CET Physics Moving Coil Galvanometer & Conversions MCQs with step-by-step solutions (20 questions). Part of Magnetic Fields due to Electric Current. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Moving Coil Galvanometer & Conversions · easy · theory
A moving coil galvanometer is an instrument used to:
A. Measure magnetic flux
B. Measure capacitance
C. Detect and measure small electric currents  ✓ Correct
D. Measure large currents directly
Solution: The current-carrying coil experiences a torque in the field of a permanent magnet, and the deflection measures the current.
Q2 — Moving Coil Galvanometer & Conversions · medium · theory
In a moving coil galvanometer, the current sensitivity is defined as:
A. The current per unit deflection
B. The deflection per unit voltage
C. The deflection produced per unit current  ✓ Correct
D. The resistance per unit deflection
Solution: Current sensitivity is $\dfrac{\theta}{I} = \dfrac{NAB}{k}$, where $k$ is the torsion constant of the suspension.
Q3 — Moving Coil Galvanometer & Conversions · easy · theory
A galvanometer is converted into an ammeter by connecting:
A. A low resistance in parallel with it  ✓ Correct
B. A high resistance in series with it
C. A capacitor in series with it
D. A high resistance in parallel with it
Solution: The shunt carries most of the current and keeps the combined resistance low, as an ammeter must have.
Q4 — Moving Coil Galvanometer & Conversions · easy · theory
A galvanometer is converted into a voltmeter by connecting:
A. An inductor in parallel with it
B. A low resistance in series with it
C. A low resistance in parallel with it
D. A high resistance in series with it  ✓ Correct
Solution: The series multiplier limits the current and gives the instrument the high resistance a voltmeter needs.
Q5 — Moving Coil Galvanometer & Conversions · medium · theory
The resistances of an ideal ammeter and an ideal voltmeter are respectively:
A. Both infinite
B. Infinite and zero
C. Both zero
D. Zero and infinite  ✓ Correct
Solution: An ideal ammeter drops no voltage in the circuit, and an ideal voltmeter draws no current from it.
Q6 — Moving Coil Galvanometer & Conversions · easy · theory
In a circuit, an ammeter and a voltmeter are connected respectively:
A. In parallel and in series with the element
B. Both in series
C. Both in parallel
D. In series and in parallel with the element  ✓ Correct
Solution: An ammeter must carry the current being measured; a voltmeter must experience the potential difference being measured.
Q7 — Moving Coil Galvanometer & Conversions · medium · theory
The voltage sensitivity of a galvanometer is defined as:
A. The potential difference per unit deflection
B. The current per unit potential difference
C. The deflection produced per unit potential difference  ✓ Correct
D. The deflection per unit current
Solution: Voltage sensitivity equals current sensitivity divided by the galvanometer resistance.
Q8 — Moving Coil Galvanometer & Conversions · hard · theory
Increasing the current sensitivity of a galvanometer does not necessarily increase its voltage sensitivity because:
A. The deflection becomes independent of current
B. The magnetic field becomes weaker
C. The change may also increase the coil resistance  ✓ Correct
D. The torsion constant becomes zero
Solution: Adding turns raises $NAB$ but also lengthens the coil, raising $G$; voltage sensitivity $\dfrac{NAB}{kG}$ may then be unchanged.
Q9 — Moving Coil Galvanometer & Conversions · hard · numerical
A galvanometer of resistance $99\,\Omega$ gives full-scale deflection at $10\text{ mA}$. To convert it into an ammeter of range $0$–$1\text{ A}$, the shunt required is:
A. $0.1\,\Omega$
B. $0.99\,\Omega$
C. $1\,\Omega$  ✓ Correct
D. $10\,\Omega$
Solution: $S = \dfrac{I_gG}{I - I_g} = \dfrac{0.01 \times 99}{1 - 0.01} = \dfrac{0.99}{0.99} = 1\,\Omega$.
Q10 — Moving Coil Galvanometer & Conversions · hard · numerical
A galvanometer of resistance $50\,\Omega$ gives full-scale deflection at $2\text{ mA}$. To convert it into a voltmeter of range $0$–$10\text{ V}$, the series resistance required is:
A. $4500\,\Omega$
B. $5000\,\Omega$
C. $4950\,\Omega$  ✓ Correct
D. $5050\,\Omega$
Solution: $R_s = \dfrac{V}{I_g} - G = \dfrac{10}{0.002} - 50 = 5000 - 50 = 4950\,\Omega$.
Q11 — Moving Coil Galvanometer & Conversions · hard · numerical
A galvanometer of resistance $100\,\Omega$ gives full-scale deflection at $1\text{ mA}$. The shunt needed to convert it into an ammeter of range $0$–$1\text{ A}$ is approximately:
A. $1\,\Omega$
B. $10\,\Omega$
C. $0.1\,\Omega$  ✓ Correct
D. $0.01\,\Omega$
Solution: $S = \dfrac{0.001 \times 100}{1 - 0.001} = \dfrac{0.1}{0.999} \approx 0.1\,\Omega$.
Q12 — Moving Coil Galvanometer & Conversions · hard · numerical
A galvanometer of resistance $20\,\Omega$ gives full-scale deflection at $5\text{ mA}$. To read up to $5\text{ V}$, the series resistance required is:
A. $900\,\Omega$
B. $1020\,\Omega$
C. $980\,\Omega$  ✓ Correct
D. $1000\,\Omega$
Solution: $R_s = \dfrac{5}{0.005} - 20 = 1000 - 20 = 980\,\Omega$.
Q13 — Moving Coil Galvanometer & Conversions · hard · numerical
A galvanometer of resistance $10\,\Omega$ gives full-scale deflection at $10\text{ mA}$. The shunt needed for a range of $0$–$0.1\text{ A}$ is approximately:
A. $1.11\,\Omega$  ✓ Correct
B. $10\,\Omega$
C. $0.1\,\Omega$
D. $0.9\,\Omega$
Solution: $S = \dfrac{0.01 \times 10}{0.1 - 0.01} = \dfrac{0.1}{0.09} \approx 1.11\,\Omega$.
Q14 — Moving Coil Galvanometer & Conversions · medium · numerical
A galvanometer of resistance $60\,\Omega$ gives full-scale deflection at $1\text{ mA}$. To read up to $3\text{ V}$, the series resistance required is:
A. $2900\,\Omega$
B. $3060\,\Omega$
C. $2940\,\Omega$  ✓ Correct
D. $3000\,\Omega$
Solution: $R_s = \dfrac{3}{0.001} - 60 = 3000 - 60 = 2940\,\Omega$.
Q15 — Moving Coil Galvanometer & Conversions · medium · numerical
The shunt resistance required to convert a galvanometer of resistance $G$ and full-scale current $I_g$ into an ammeter of range $I$ is:
A. $\dfrac{IG}{I_g}$
B. $\dfrac{I_gG}{I - I_g}$  ✓ Correct
C. $\dfrac{G}{I - I_g}$
D. $\dfrac{I - I_g}{I_gG}$
Solution: The shunt must carry $I - I_g$ while the same potential difference appears across both branches.
Q16 — Moving Coil Galvanometer & Conversions · medium · numerical
A galvanometer of resistance $40\,\Omega$ gives full-scale deflection at $5\text{ mA}$. To read up to $2\text{ V}$, the series resistance required is:
A. $360\,\Omega$  ✓ Correct
B. $440\,\Omega$
C. $340\,\Omega$
D. $400\,\Omega$
Solution: $R_s = \dfrac{2}{0.005} - 40 = 400 - 40 = 360\,\Omega$.
Q17 — Moving Coil Galvanometer & Conversions · medium · numerical
A galvanometer shows a deflection of $30$ divisions for a current of $2\text{ mA}$. Its current sensitivity is:
A. $15$ divisions per milliampere  ✓ Correct
B. $0.067$ divisions per milliampere
C. $30$ divisions per milliampere
D. $60$ divisions per milliampere
Solution: Current sensitivity $= \dfrac{30}{2} = 15$ divisions per milliampere.
Q18 — Moving Coil Galvanometer & Conversions · hard · numerical
A galvanometer of resistance $99\,\Omega$ gives full-scale deflection at $1\text{ mA}$. The shunt needed for a range of $0$–$0.1\text{ A}$ is:
A. $1\,\Omega$  ✓ Correct
B. $0.1\,\Omega$
C. $9.9\,\Omega$
D. $0.99\,\Omega$
Solution: $S = \dfrac{0.001 \times 99}{0.1 - 0.001} = \dfrac{0.099}{0.099} = 1\,\Omega$.
Q19 — Moving Coil Galvanometer & Conversions · medium · numerical
When a shunt is connected across a galvanometer, the resistance of the resulting ammeter is:
A. Less than both the galvanometer resistance and the shunt resistance  ✓ Correct
B. Greater than the galvanometer resistance
C. Equal to the sum of the two
D. Equal to the shunt resistance
Solution: The two are in parallel, and a parallel combination is always smaller than either branch.
Q20 — Moving Coil Galvanometer & Conversions · hard · numerical
To double the range of a voltmeter made from a galvanometer, the required series resistance must be:
A. Left unchanged
B. More than doubled, since the galvanometer resistance is subtracted only once  ✓ Correct
C. Reduced to one-fourth
D. Exactly halved
Solution: From $R_s = \dfrac{V}{I_g} - G$, doubling $V$ doubles the first term but $G$ is still subtracted once, so $R_s$ becomes slightly more than twice as large.