201. A moving coil galvanometer works mainly on which principle?
ⓐ. A stationary charge placed in a magnetic field experiences force
ⓑ. A resistor in parallel always increases voltage
ⓒ. A solenoid has exactly zero field inside it
ⓓ. A current-carrying coil in a magnetic field experiences torque
Correct Answer: A current-carrying coil in a magnetic field experiences torque
Explanation: A moving coil galvanometer contains a coil suspended in a magnetic field. When current passes through the coil, the two sides of the coil experience magnetic forces. These forces form a couple and produce torque on the coil. The coil rotates until the magnetic torque is balanced by the restoring torque of the suspension or spring. Thus the deflection of the coil becomes a measure of current. The device is not based on force on a stationary charge; current in the coil is essential.
202. Why is a radial magnetic field used in a moving coil galvanometer?
ⓐ. To keep torque proportional to current and produce a uniform scale
ⓑ. To make the magnetic field zero inside the coil at every deflection
ⓒ. To make the restoring torque independent of angular deflection
ⓓ. To make the coil resistance maximum at full-scale deflection
Correct Answer: To keep torque proportional to current and produce a uniform scale
Explanation: In a moving coil galvanometer, the magnetic torque is generally \(\tau=NIAB\sin\theta\). A radial magnetic field is arranged so that the plane of the coil remains effectively parallel to the magnetic field during rotation, making \(\sin\theta=1\) for the working range. Thus the magnetic torque becomes \(\tau=NIAB\), directly proportional to current \(I\). Since the restoring torque is \(k\phi\), equilibrium gives \(NIAB=k\phi\). This makes deflection \(\phi\) directly proportional to current. The radial field is therefore responsible for a linear or uniform scale, which is a major practical advantage.
203. At equilibrium in a moving coil galvanometer, which relation is correct?
ⓐ. \(NIA=kB\phi\)
ⓑ. \(I=\frac{NAB}{k\phi}\)
ⓒ. \(NIAB=k\phi\)
ⓓ. \(k=NIAB\phi\)
Correct Answer: \(NIAB=k\phi\)
Explanation: In a moving coil galvanometer, current in the coil produces magnetic torque. In a radial magnetic field, this torque is \(NIAB\). The suspension strip or spring produces restoring torque \(k\phi\), where \(k\) is the torsional constant and \(\phi\) is angular deflection. At equilibrium, the magnetic torque is balanced by the restoring torque. Therefore, \(NIAB=k\phi\). This relation also shows that \(\phi\propto I\) when \(N\), \(A\), \(B\), and \(k\) are constant. A common algebra-related mistake is to put \(\phi\) in the denominator before applying the torque balance.
204. A moving coil galvanometer has \(N=100\), \(A=2.0\times10^{-4}\,\text{m}^2\), \(B=0.50\,\text{T}\), and torsional constant \(k=2.0\times10^{-6}\,\text{N m rad}^{-1}\). What current produces a deflection of \(0.20\,\text{rad}\)?
ⓐ. \(2.0\times10^{-5}\,\text{A}\)
ⓑ. \(8.0\times10^{-5}\,\text{A}\)
ⓒ. \(2.0\times10^{-4}\,\text{A}\)
ⓓ. \(4.0\times10^{-5}\,\text{A}\)
Correct Answer: \(4.0\times10^{-5}\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(N=100\), \(A=2.0\times10^{-4}\,\text{m}^2\), \(B=0.50\,\text{T}\), \(k=2.0\times10^{-6}\,\text{N m rad}^{-1}\), and \(\phi=0.20\,\text{rad}\).
\( \textbf{Required:} \) Current \(I\).
\( \textbf{Equilibrium relation:} \)
\[
NIAB=k\phi
\]
\( \textbf{Solve for current:} \)
\[
I=\frac{k\phi}{NAB}
\]
\( \textbf{Substitution:} \)
\[
I=\frac{(2.0\times10^{-6})(0.20)}{(100)(2.0\times10^{-4})(0.50)}
\]
\( \textbf{Numerator:} \)
\[
(2.0\times10^{-6})(0.20)=4.0\times10^{-7}
\]
\( \textbf{Denominator:} \)
\[
(100)(2.0\times10^{-4})(0.50)=1.0\times10^{-2}
\]
\( \textbf{Final calculation:} \)
\[
I=\frac{4.0\times10^{-7}}{1.0\times10^{-2}}=4.0\times10^{-5}\,\text{A}
\]
\( \textbf{Final answer:} \) The required current is \(4.0\times10^{-5}\,\text{A}\).
205. In a moving coil galvanometer, the deflection \(\phi\) is directly proportional to current \(I\) when:
ⓐ. \(k\) increases in direct proportion to \(I\)
ⓑ. \(N\), \(A\), \(B\), and \(k\) remain constant
ⓒ. \(B\) becomes zero
ⓓ. the coil is removed from the magnetic field
Correct Answer: \(N\), \(A\), \(B\), and \(k\) remain constant
Explanation: The equilibrium relation of a moving coil galvanometer is \(NIAB=k\phi\). Rearranging gives \(\phi=\frac{NAB}{k}I\). If \(N\), \(A\), \(B\), and \(k\) remain constant, the factor \(\frac{NAB}{k}\) is constant. Therefore, the deflection is directly proportional to current. If \(B=0\), there is no magnetic torque and the galvanometer cannot work normally. The proportionality depends on the instrument parameters staying fixed; otherwise the scale would not remain uniform.
206. A graph of deflection \(\phi\) versus current \(I\) is drawn for a moving coil galvanometer in a radial magnetic field. What is the slope of the graph?
ⓐ. \(\frac{k}{NAB}\)
ⓑ. \(NIAB\)
ⓒ. \(\frac{NAB}{k}\)
ⓓ. \(\frac{I}{\phi}\)
Correct Answer: \(\frac{NAB}{k}\)
Explanation: \( \textbf{Torque balance:} \)
\[
NIAB=k\phi
\]
\( \textbf{Rearrange for deflection:} \)
\[
\phi=\frac{NAB}{k}I
\]
\( \textbf{Graph form:} \) This has the form \(y=mx\), where \(\phi\) is on the vertical axis and \(I\) is on the horizontal axis.
\( \textbf{Slope:} \)
\[
\text{slope}=\frac{\phi}{I}=\frac{NAB}{k}
\]
\( \textbf{Meaning:} \) The slope is the current sensitivity of the galvanometer.
\( \textbf{Common mistake:} \) \(\frac{k}{NAB}\) is the current required per unit deflection, not the slope of \(\phi\) versus \(I\).
\( \textbf{Final answer:} \) The slope is \(\frac{NAB}{k}\).
207. Which change increases the current sensitivity of a moving coil galvanometer, assuming other factors remain unchanged?
ⓐ. Increasing \(N\), \(A\), or \(B\)
ⓑ. Increasing torsional constant \(k\)
ⓒ. Decreasing magnetic field \(B\)
ⓓ. Reducing the number of turns \(N\)
Correct Answer: Increasing \(N\), \(A\), or \(B\)
Explanation: Current sensitivity is deflection per unit current:
\[
\frac{\phi}{I}=\frac{NAB}{k}
\]
This expression shows that current sensitivity increases when \(N\), \(A\), or \(B\) increases. It decreases when the torsional constant \(k\) increases because a stiffer suspension gives smaller deflection for the same current. Reducing \(N\) or reducing \(B\) weakens magnetic torque and reduces sensitivity. The important instrument-design idea is that a sensitive galvanometer should give a larger deflection for a smaller current. The most tempting wrong option is increasing \(k\), but that makes the spring harder to twist.
208. Assertion: A radial magnetic field gives a moving coil galvanometer a uniform scale.
Reason: In a radial field, magnetic torque remains proportional to current without a changing \(\sin\theta\) factor during deflection.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A uniform scale means equal increases in current produce equal increases in deflection. In a moving coil galvanometer, the magnetic torque should therefore remain directly proportional to current. A radial magnetic field keeps the effective angle condition such that the torque is \(NIAB\), rather than \(NIAB\sin\theta\) with changing \(\theta\). At equilibrium, \(NIAB=k\phi\), so \(\phi\propto I\). The Reason correctly explains the Assertion because removal of the changing sine factor is what makes the scale linear. Without a radial field, the torque-deflection relation would not remain as simply proportional over the working range.
209. In a moving coil galvanometer, current sensitivity is defined as:
ⓐ. \(\frac{\phi}{I}\)
ⓑ. \(\frac{I}{\phi}\)
ⓒ. \(\frac{\phi}{V}\)
ⓓ. \(\frac{V}{\phi}\)
Correct Answer: \(\frac{\phi}{I}\)
Explanation: Current sensitivity of a galvanometer means the deflection produced per unit current. If a current \(I\) produces deflection \(\phi\), then current sensitivity is \(\frac{\phi}{I}\). From the galvanometer relation \(NIAB=k\phi\), we get \(\frac{\phi}{I}=\frac{NAB}{k}\). A larger value of \(\frac{\phi}{I}\) means a smaller current can produce a measurable deflection. The reciprocal \(\frac{I}{\phi}\) tells current needed per unit deflection, not sensitivity. The important distinction is between current sensitivity and voltage sensitivity, which is deflection per unit voltage.
210. The voltage sensitivity of a galvanometer of resistance \(G\) is:
ⓐ. \(\frac{\phi}{V}=\frac{kG}{NAB}\)
ⓑ. \(\frac{\phi}{V}=\frac{NAB}{kG}\)
ⓒ. \(\frac{\phi}{V}=\frac{NABG}{k}\)
ⓓ. \(\frac{\phi}{V}=\frac{k}{NABG}\)
Correct Answer: \(\frac{\phi}{V}=\frac{NAB}{kG}\)
Explanation: \( \textbf{Current-deflection relation:} \)
\[
\phi=\frac{NAB}{k}I
\]
\( \textbf{Voltage across galvanometer:} \) If the galvanometer resistance is \(G\), then
\[
V=IG
\]
\( \textbf{Current in terms of voltage:} \)
\[
I=\frac{V}{G}
\]
\( \textbf{Substitute in deflection relation:} \)
\[
\phi=\frac{NAB}{k}\cdot\frac{V}{G}
\]
\( \textbf{Voltage sensitivity:} \)
\[
\frac{\phi}{V}=\frac{NAB}{kG}
\]
\( \textbf{Meaning:} \) Voltage sensitivity decreases if \(G\) increases while \(N\), \(A\), \(B\), and \(k\) remain fixed.
\( \textbf{Final answer:} \) \(\frac{\phi}{V}=\frac{NAB}{kG}\).
211. If the galvanometer resistance \(G\) is increased while \(N\), \(A\), \(B\), and \(k\) remain unchanged, what happens to current sensitivity and voltage sensitivity?
ⓐ. Current sensitivity decreases, voltage sensitivity remains unchanged
ⓑ. Both current sensitivity and voltage sensitivity increase
ⓒ. Both current sensitivity and voltage sensitivity remain unchanged
ⓓ. Current sensitivity remains unchanged, voltage sensitivity decreases
Correct Answer: Current sensitivity remains unchanged, voltage sensitivity decreases
Explanation: Current sensitivity is \(\frac{\phi}{I}=\frac{NAB}{k}\). This expression does not contain galvanometer resistance \(G\), so changing \(G\) alone does not change current sensitivity. Voltage sensitivity is \(\frac{\phi}{V}=\frac{NAB}{kG}\). This expression contains \(G\) in the denominator. Therefore, increasing \(G\) decreases voltage sensitivity if the other parameters remain fixed. This is why increasing current sensitivity does not automatically guarantee increased voltage sensitivity. A common mistake is to assume that every improvement in deflection per current also improves deflection per voltage.
212. A galvanometer has \(N=200\), \(A=1.5\times10^{-4}\,\text{m}^2\), \(B=0.40\,\text{T}\), \(k=3.0\times10^{-6}\,\text{N m rad}^{-1}\), and resistance \(G=60\,\Omega\). What is its voltage sensitivity?
ⓐ. \(66.7\,\text{rad V}^{-1}\)
ⓑ. \(33.3\,\text{rad V}^{-1}\)
ⓒ. \(0.033\,\text{rad V}^{-1}\)
ⓓ. \(0.067\,\text{rad V}^{-1}\)
Correct Answer: \(66.7\,\text{rad V}^{-1}\)
Explanation: \( \textbf{Given:} \) \(N=200\), \(A=1.5\times10^{-4}\,\text{m}^2\), \(B=0.40\,\text{T}\), \(k=3.0\times10^{-6}\,\text{N m rad}^{-1}\), and \(G=60\,\Omega\).
\( \textbf{Required:} \) Voltage sensitivity \(\frac{\phi}{V}\).
\( \textbf{Formula:} \)
\[
\frac{\phi}{V}=\frac{NAB}{kG}
\]
\( \textbf{Numerator:} \)
\[
NAB=(200)(1.5\times10^{-4})(0.40)
\]
\[
(200)(1.5\times10^{-4})=3.0\times10^{-2}
\]
\[
(3.0\times10^{-2})(0.40)=1.2\times10^{-2}
\]
\( \textbf{Denominator:} \)
\[
kG=(3.0\times10^{-6})(60)=1.8\times10^{-4}
\]
\( \textbf{Final calculation:} \)
\[
\frac{\phi}{V}=\frac{1.2\times10^{-2}}{1.8\times10^{-4}}=66.7\,\text{rad V}^{-1}
\]
\( \textbf{Final answer:} \) The voltage sensitivity is \(66.7\,\text{rad V}^{-1}\).
213. Which row correctly compares current sensitivity and voltage sensitivity of a galvanometer?
| Row | Quantity | Expression | Depends on \(G\)? |
|---|
| P | Current sensitivity | \(\frac{\phi}{I}=\frac{NAB}{k}\) | No, if \(N\), \(A\), \(B\), \(k\) are fixed |
| Q | Voltage sensitivity | \(\frac{\phi}{V}=\frac{NABG}{k}\) | Directly proportional to \(G\) |
| R | Current sensitivity | \(\frac{\phi}{I}=\frac{k}{NAB}\) | Directly proportional to \(G\) |
| S | Voltage sensitivity | \(\frac{\phi}{V}=\frac{kG}{NAB}\) | Independent of \(G\) |
ⓐ. Row Q
ⓑ. Row P
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Current sensitivity is \(\frac{\phi}{I}=\frac{NAB}{k}\), so row P is correct. Voltage sensitivity is \(\frac{\phi}{V}=\frac{NAB}{kG}\), not \(\frac{NABG}{k}\), so row Q is incorrect. Row R gives the reciprocal of current sensitivity and wrongly links it directly with \(G\). Row S gives the reciprocal-style wrong expression for voltage sensitivity and also wrongly says it is independent of \(G\). The main distinction is that \(G\) enters when voltage is involved because \(V=IG\). Current sensitivity is about deflection per current, while voltage sensitivity is about deflection per applied voltage.
214. Assertion: Increasing the number of turns \(N\) of a galvanometer coil may increase current sensitivity.
Reason: Current sensitivity is \(\frac{\phi}{I}=\frac{NAB}{k}\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Current sensitivity is the deflection produced per unit current. From the galvanometer torque balance, \(NIAB=k\phi\), so \(\frac{\phi}{I}=\frac{NAB}{k}\). This expression shows that increasing \(N\) increases current sensitivity if \(A\), \(B\), and \(k\) remain unchanged. The Reason gives the exact mathematical basis for the Assertion. In real instruments, increasing turns may also change resistance and practical design features, but the direct sensitivity formula supports the stated result. A common mistake is to mix current sensitivity with voltage sensitivity, where resistance \(G\) also matters.
215. A galvanometer has current sensitivity \(5.0\times10^4\,\text{rad A}^{-1}\) and resistance \(100\,\Omega\). What is its voltage sensitivity?
ⓐ. \(5.0\times10^6\,\text{rad V}^{-1}\)
ⓑ. \(5.0\times10^4\,\text{rad V}^{-1}\)
ⓒ. \(5.0\times10^2\,\text{rad V}^{-1}\)
ⓓ. \(5.0\times10^{-2}\,\text{rad V}^{-1}\)
Correct Answer: \(5.0\times10^2\,\text{rad V}^{-1}\)
Explanation: \( \textbf{Given:} \) Current sensitivity \(S_I=\frac{\phi}{I}=5.0\times10^4\,\text{rad A}^{-1}\), and galvanometer resistance \(G=100\,\Omega\).
\( \textbf{Required:} \) Voltage sensitivity \(S_V=\frac{\phi}{V}\).
\( \textbf{Voltage-current relation:} \)
\[
V=IG
\]
\( \textbf{Convert sensitivity:} \)
\[
S_V=\frac{\phi}{V}=\frac{\phi}{IG}
\]
\( \textbf{Use current sensitivity:} \)
\[
S_V=\frac{1}{G}\left(\frac{\phi}{I}\right)
\]
\( \textbf{Substitution:} \)
\[
S_V=\frac{5.0\times10^4}{100}
\]
\( \textbf{Final simplification:} \)
\[
S_V=5.0\times10^2\,\text{rad V}^{-1}
\]
\( \textbf{Final answer:} \) The voltage sensitivity is \(5.0\times10^2\,\text{rad V}^{-1}\).
216. To convert a galvanometer into an ammeter of higher range, a suitable resistance is connected:
ⓐ. In series with the galvanometer and having high resistance
ⓑ. In series with the galvanometer and having zero current
ⓒ. In parallel with the galvanometer and having low resistance
ⓓ. In parallel with the galvanometer and having infinite resistance
Correct Answer: In parallel with the galvanometer and having low resistance
Explanation: A galvanometer can safely carry only a small current \(I_g\). To convert it into an ammeter, most of the total current must bypass the galvanometer. This is done by connecting a low resistance called a shunt in parallel with the galvanometer. Since parallel branches have the same potential difference, the low-resistance shunt carries the larger part of the current. The resulting instrument has low equivalent resistance, which is suitable for series connection in a circuit. The common mistake is to use a high series resistance, but that is used for converting a galvanometer into a voltmeter, not an ammeter.
217. A galvanometer of resistance \(G\) gives full-scale deflection at current \(I_g\). It is converted into an ammeter of range \(I\) by connecting shunt resistance \(S\). Which expression is correct?
ⓐ. \(S=\frac{IG}{I_g-I}\)
ⓑ. \(S=\frac{(I-I_g)G}{I_g}\)
ⓒ. \(S=\frac{I_g}{G(I-I_g)}\)
ⓓ. \(S=\frac{I_gG}{I-I_g}\)
Correct Answer: \(S=\frac{I_gG}{I-I_g}\)
Explanation: \( \textbf{Ammeter conversion condition:} \) The galvanometer and shunt are connected in parallel.
\( \textbf{Galvanometer current:} \) At full scale, galvanometer current is \(I_g\).
\( \textbf{Shunt current:} \)
\[
I_s=I-I_g
\]
\( \textbf{Same potential difference:} \) Since the galvanometer and shunt are parallel,
\[
I_gG=I_sS
\]
\( \textbf{Substitute \(I_s\):} \)
\[
I_gG=(I-I_g)S
\]
\( \textbf{Solve for shunt:} \)
\[
S=\frac{I_gG}{I-I_g}
\]
\( \textbf{Condition check:} \) Since \(I\) is much larger than \(I_g\), \(S\) is usually much smaller than \(G\).
\( \textbf{Final answer:} \) \(S=\frac{I_gG}{I-I_g}\).
218. A galvanometer has resistance \(G=99\,\Omega\) and full-scale current \(I_g=1.0\,\text{mA}\). What shunt resistance is needed to convert it into an ammeter of range \(1.0\,\text{A}\)?
ⓐ. \(9.9\times10^{-2}\,\Omega\)
ⓑ. \(9.9\times10^{-1}\,\Omega\)
ⓒ. \(9.9\times10^{0}\,\Omega\)
ⓓ. \(9.9\times10^{1}\,\Omega\)
Correct Answer: \(9.9\times10^{-2}\,\Omega\)
Explanation: \( \textbf{Given:} \) \(G=99\,\Omega\), \(I_g=1.0\,\text{mA}=1.0\times10^{-3}\,\text{A}\), and \(I=1.0\,\text{A}\).
\( \textbf{Required:} \) Shunt resistance \(S\).
\( \textbf{Formula:} \)
\[
S=\frac{I_gG}{I-I_g}
\]
\( \textbf{Shunt current denominator:} \)
\[
I-I_g=1.0-0.001=0.999\,\text{A}
\]
\( \textbf{Numerator:} \)
\[
I_gG=(1.0\times10^{-3})(99)=0.099
\]
\( \textbf{Substitution:} \)
\[
S=\frac{0.099}{0.999}
\]
\( \textbf{Simplification:} \)
\[
S\approx0.099\,\Omega
\]
\( \textbf{Physical check:} \) The shunt is much smaller than \(G\), so most current bypasses the galvanometer.
\( \textbf{Final answer:} \) The required shunt resistance is approximately \(0.099\,\Omega\).
219. A galvanometer of resistance \(50\,\Omega\) is shunted by a resistance \(0.50\,\Omega\). If the galvanometer current is \(2.0\,\text{mA}\), what is the current through the shunt?
ⓐ. \(0.020\,\text{A}\)
ⓑ. \(2.0\,\text{A}\)
ⓒ. \(20\,\text{A}\)
ⓓ. \(0.20\,\text{A}\)
Correct Answer: \(0.20\,\text{A}\)
Explanation: \( \textbf{Given:} \) \(G=50\,\Omega\), \(S=0.50\,\Omega\), and \(I_g=2.0\,\text{mA}=2.0\times10^{-3}\,\text{A}\).
\( \textbf{Parallel condition:} \) The galvanometer and shunt have the same potential difference.
\( \textbf{Voltage across galvanometer:} \)
\[
V_g=I_gG
\]
\( \textbf{Substitution:} \)
\[
V_g=(2.0\times10^{-3})(50)=0.10\,\text{V}
\]
\( \textbf{Voltage across shunt:} \)
\[
V_s=V_g=0.10\,\text{V}
\]
\( \textbf{Shunt current:} \)
\[
I_s=\frac{V_s}{S}
\]
\( \textbf{Substitution:} \)
\[
I_s=\frac{0.10}{0.50}=0.20\,\text{A}
\]
\( \textbf{Current division check:} \) The shunt has much smaller resistance, so it carries much larger current than the galvanometer.
\( \textbf{Final answer:} \) The shunt current is \(0.20\,\text{A}\).
220. Why is an ideal ammeter expected to have very low resistance?
ⓐ. So that no current can pass through it
ⓑ. So it negligibly changes the current being measured
ⓒ. So that it must always be connected in parallel with the load
ⓓ. So that it increases the circuit resistance as much as possible
Correct Answer: So it negligibly changes the current being measured
Explanation: An ammeter is connected in series with the circuit element whose current is to be measured. If the ammeter had high resistance, it would significantly increase the total circuit resistance. That would reduce the current and disturb the quantity being measured. Therefore, an ideal ammeter should have zero or very low resistance. The shunt used in ammeter conversion helps reduce the effective resistance of the instrument. The measurement-related mistake is to connect an ammeter in parallel like a voltmeter; an ammeter is a series instrument and must not appreciably oppose the circuit current.
