Magnetic Effects of Current and Magnetism MCQs with Answers – Part 5 (NEET Physics)

Explanation: The magnitude of the magnetic force per unit length between long parallel wires is \[ f=\frac{\mu_0I_1I_2}{2\pi d} \] where the current magnitudes are used in the force-magnitude calculation. For equal currents, \[ f=\frac{\mu_0I^2}{2\pi d} \] The traditional force-based reference uses equal currents of \(1\,\text{A}\), with the wires separated by \(1\,\text{m}\) in vacuum. Its characteristic force per unit length is approximately \[ f_{\mathrm{ref}}=2\times10^{-7}\,\text{N m}^{-1} \] If a different pair of equal currents reproduces this force magnitude, its current–separation combination must satisfy the corresponding proportionality. The equality of force magnitude alone does not uniquely establish either current. Statement I is incorrect. For \(I=3\,\text{A}\), the product of the two current magnitudes is nine times the \(1\,\text{A}\) reference product. A separation nine times as large compensates for that change. Therefore, the required separation is \(9\,\text{m}\), making Statement II correct. The force-magnitude formula does not distinguish attraction from repulsion. That distinction follows from the relative current directions. Equal-direction currents attract, while opposite-direction currents repel. Consequently, a repulsive interaction can have the same force magnitude per unit length as an attractive reference interaction. Statement III is correct. The traditional reference also specifies the geometry, idealised conductor dimensions, vacuum conditions and equal specified currents. Matching only one measured force magnitude does not establish all those conditions. Statement IV is correct. The force-based description is the historical treatment relevant to this concept. It should not be represented as the current formal SI definition of the ampere, which is based on the fixed value of the elementary charge.

Explanation: \(\textbf{Establish the expected force on a measured length:}\) For sufficiently long parallel wires, \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] Therefore, the force magnitude on a specified observation length \(L\) is \[ F=\frac{\mu_0I_1I_2L}{2\pi d} \] The measured force must scale with the current product, the inverse separation and the observation length. \(\textbf{Check group P:}\) Substituting its values, \[ F_P=(2.0\times10^{-7})\frac{(1)(1)(0.50)}{1} \] \[ F_P=1.0\times10^{-7}\,\text{N} \] This agrees with the reported value. \(\textbf{Check group Q:}\) \[ F_Q=(2.0\times10^{-7})\frac{(2)(1)(1.0)}{2} \] \[ F_Q=2.0\times10^{-7}\,\text{N} \] This also agrees with its reported value. \(\textbf{Check group R:}\) \[ F_R=(2.0\times10^{-7})\frac{(3)(1)(0.25)}{1.5} \] \[ F_R=1.0\times10^{-7}\,\text{N} \] This matches the reported value. \(\textbf{Check group S:}\) \[ F_S=(2.0\times10^{-7})\frac{(2)(2)(0.50)}{2} \] \[ F_S=2.0\times10^{-7}\,\text{N} \] However, group S reports only \(1.0\times10^{-7}\,\text{N}\). Its reported force is half the predicted value. Therefore, S alone is inconsistent with the stated ideal interaction model. The analysis requires using the specified measured lengths rather than assuming that every group reports force per unit length. Different combinations of currents, separation and length can produce equal total force magnitudes. A report cannot be rejected merely because its conductor currents differ from the traditional \(1\,\text{A}\) reference arrangement. The inconsistency is established only after evaluating the complete force relation.

Explanation: The force per unit length between two sufficiently long, straight, parallel current-carrying conductors is \[ \frac FL=\frac{\mu_0I_1I_2}{2\pi d} \] The left-hand side has units of newtons per metre. The traditional reference uses the characteristic force density of approximately \[ 2\times10^{-7}\,\text{N m}^{-1} \] A total force in newtons is not sufficient without also specifying the corresponding conductor length. Statement I is correct. The reference arrangement specifies straight parallel conductors, sufficiently great length, negligible transverse dimensions, a separation of \(1\,\text{m}\), and vacuum conditions. These qualifications connect the reference to the ideal long-wire force law. Statement II is correct. The force relation depends on the product \(I_1I_2\) divided by separation \(d\). Different current pairs and separations can therefore produce the same force density. A numerical equality with the reference force does not prove that each current is \(1\,\text{A}\). Statement III is incorrect. The historical reference specifically uses two equal currents of \(1\,\text{A}\) under its stated conditions. The current formal SI definition is instead based on fixing the numerical value of the elementary charge when expressed in coulombs. The historical mechanical relation remains useful for studying current interactions, but it is not itself the present definition. Statement IV is correct. The distinction matters when interpreting a modern experimental measurement. A force calculation may reproduce the historical benchmark without establishing the current unit through that measurement. Correct interpretation requires separating the physical interaction law from the metrological definition of the unit.

Explanation: \(\textbf{Write the original interaction relation:}\) For two sufficiently long parallel conductors, \[ f=\frac{\mu_0I^2}{2\pi d} \] Both currents initially have the same direction. The interaction is therefore attractive. \(\textbf{Apply the changed current magnitudes:}\) Each current is doubled: \[ I'_1=2I,\qquad I'_2=2I \] The new product of current magnitudes is \[ I'_1I'_2=(2I)(2I) \] \[ I'_1I'_2=4I^2 \] Thus, changing both current magnitudes contributes a factor of four. \(\textbf{Apply the changed separation:}\) The new separation is \[ d'=3d \] The force per unit length varies inversely with separation, not with its square. Therefore, \[ f'=\frac{\mu_0(4I^2)}{2\pi(3d)} \] \[ f'=\frac43\frac{\mu_0I^2}{2\pi d} \] \[ f'=\frac43f \] \(\textbf{Determine the changed interaction direction:}\) Reversing only one current makes the two currents opposite in direction. The interaction therefore changes from attraction to repulsion. The reversal does not introduce an additional factor into the force magnitude because the current magnitudes are unchanged by reversal. \(\textbf{Identify the student's incorrect dependence:}\) The student's factor \(4/9\) corresponds to dividing the current-product factor of four by the square of the separation factor of three. That would be appropriate for an inverse-square separation law, but the parallel-wire interaction follows \[ f\propto\frac{I_1I_2}{d} \] \(\textbf{Final answer:}\) \[ \boxed{f'=\frac43f\ \text{(repulsive)}} \] The long-parallel-wire force relation contains the product of the two currents and one inverse power of separation. Both current magnitudes must be included when calculating the new force density. The current reversal controls the interaction direction separately from its magnitude. The modification increases the force per unit length despite increasing the separation, because the current-product increase is larger.

Explanation: The force-density magnitude for two long parallel wires is \[ f=\frac{\mu_0I^2}{2\pi d} \] Using the stated reference, \[ f=f_{\mathrm{ref}}\frac{I^2/(1\,\text{A})^2}{d/(1\,\text{m})} \] The numerical factor therefore depends on the square of the equal current magnitude divided by separation in the given units. The interaction sign depends separately on the current directions. Same-direction currents attract and receive a positive sign under the question's convention. Opposite-direction currents repel and receive a negative sign. \(\textbf{Evaluate case P:}\) The current and separation match the traditional reference arrangement: \[ f_P=+f_{\mathrm{ref}} \] \(\textbf{Evaluate case Q:}\) The current factor is \(2^2=4\), while the separation factor is four. Therefore, \[ f_Q=-\frac44f_{\mathrm{ref}} \] \[ f_Q=-f_{\mathrm{ref}} \] \(\textbf{Evaluate case R:}\) The current factor is \(3^2=9\), while the separation factor is three. Thus, \[ f_R=+\frac93f_{\mathrm{ref}} \] \[ f_R=+3f_{\mathrm{ref}} \] \(\textbf{Evaluate case S:}\) The current factor is four, while the separation factor is two. The opposite current directions make the interaction repulsive: \[ f_S=-\frac42f_{\mathrm{ref}} \] \[ f_S=-2f_{\mathrm{ref}} \] \(\textbf{Arrange the signed force densities:}\) The values are \[ f_S=-2f_{\mathrm{ref}},\quad f_Q=-f_{\mathrm{ref}},\quad f_P=+f_{\mathrm{ref}},\quad f_R=+3f_{\mathrm{ref}} \] Therefore, \[ f_S\lt f_Q\lt f_P\lt f_R \] The required order is S, Q, P, R. A sign convention for attraction and repulsion must not be confused with a direction along a particular Cartesian axis. The signed values here distinguish the two kinds of interaction without specifying wire positions. The force magnitudes follow the current-product and separation relation. Case Q reproduces the traditional force magnitude but not its equal-current reference geometry. The numerical benchmark alone is therefore insufficient to identify the traditional reference arrangement.

Explanation: \(\textbf{Establish the full-scale voltage condition:}\) A galvanometer is converted into a voltmeter by connecting a large resistance in series with its moving coil. The resulting combination is connected in parallel across the potential difference to be measured. Let the galvanometer resistance be \(G\) and the added series resistance be \(R\). Both resistances carry the same current. At the desired full-scale voltage \(V_g\), the moving coil must carry its full-scale current \(I_g\). Therefore, \[ V_g=I_g(G+R) \] \(\textbf{Determine the multiplier resistance:}\) Rearranging, \[ R=\frac{V_g}{I_g}-G \] Substituting, \[ R=\frac{10}{0.50\times10^{-3}}-100 \] \[ R=20000-100 \] \[ R=19900\,\Omega \] \[ R=19.9\,\text{k}\Omega \] \(\textbf{Calculate the current at the applied voltage:}\) The effective voltmeter resistance is \[ R_V=G+R \] \[ R_V=100+19900 \] \[ R_V=20000\,\Omega \] With an ideal \(8.0\,\text{V}\) source, \[ I=\frac{V}{R_V} \] \[ I=\frac{8.0}{20000} \] \[ I=0.40\,\text{mA} \] \(\textbf{Determine the angular deflection:}\) The moving-coil galvanometer operates linearly in its radial magnetic field. Therefore, \[ \frac{\phi}{\phi_g}=\frac{I}{I_g} \] Substituting, \[ \phi=0.40\left(\frac{0.40}{0.50}\right) \] \[ \phi=0.32\,\text{rad} \] \(\textbf{Check the range condition:}\) The applied voltage is \(80\%\) of full scale. The current and deflection are also \(80\%\) of their full-scale values. Thus, the moving coil operates within its stated range. \(\textbf{Final answer:}\) \[ \boxed{R=19.9\,\text{k}\Omega,\qquad I=0.40\,\text{mA},\qquad \phi=0.32\,\text{rad}} \] The added resistance limits the current through the sensitive galvanometer. The conversion changes the voltage range, not the intrinsic full-scale current or current sensitivity of its moving coil. The galvanometer resistance must be subtracted from the required total resistance. Using the total resistance as the external multiplier would produce a slightly incorrect full-scale calibration.

Explanation: \(\textbf{Determine the resistance required on the lower range:}\) The full-scale current is \[ I_g=1.0\times10^{-3}\,\text{A} \] On the \(10\,\text{V}\) range, the galvanometer and \(R_1\) are in series. Therefore, \[ 10=I_g(G+R_1) \] Rearranging, \[ R_1=\frac{10}{I_g}-G \] Substituting, \[ R_1=\frac{10}{1.0\times10^{-3}}-200 \] \[ R_1=10000-200 \] \[ R_1=9800\,\Omega \] \(\textbf{Determine the total resistance on the higher range:}\) The full-scale voltage is now \(50\,\text{V}\). The same galvanometer full-scale current must be maintained. Therefore, \[ 50=I_g(G+R_1+R_2) \] The required total resistance is \[ G+R_1+R_2=\frac{50}{1.0\times10^{-3}} \] \[ G+R_1+R_2=50000\,\Omega \] The galvanometer and first resistor already contribute \[ G+R_1=200+9800 \] \[ G+R_1=10000\,\Omega \] Hence, \[ R_2=50000-10000 \] \[ R_2=40000\,\Omega \] \(\textbf{Determine the coil voltage at the applied }30\,\text{V}\textbf{:}\) The instrument is set to the \(50\,\text{V}\) range. Its effective resistance is therefore \(50000\,\Omega\). The current through every series element is \[ I=\frac{30}{50000} \] \[ I=0.60\,\text{mA} \] The voltage across the galvanometer coil is \[ V_G=IG \] \[ V_G=(0.60\times10^{-3})(200) \] \[ V_G=0.12\,\text{V} \] \(\textbf{Final answer:}\) \[ \boxed{R_1=9.8\,\text{k}\Omega,\qquad R_2=40.0\,\text{k}\Omega,\qquad V_G=0.12\,\text{V}} \] The higher voltage range requires a greater total series resistance because the galvanometer's permissible current remains unchanged. The additional resistor \(R_2\) extends the existing \(10\,\text{V}\) arrangement to \(50\,\text{V}\). It must not be calculated as though the lower-range resistor were absent. Although \(30\,\text{V}\) appears across the complete voltmeter, only a small fraction appears across the sensitive moving coil.

Explanation: \(\textbf{Recover the intended total resistance:}\) A voltmeter reaches full-scale deflection when the galvanometer carries its specified full-scale current. For the intended \(10\,\text{V}\) range, \[ R_{V,\mathrm{intended}}=\frac{V_g}{I_g} \] Substituting, \[ R_{V,\mathrm{intended}}=\frac{10}{1.0\times10^{-3}} \] \[ R_{V,\mathrm{intended}}=10000\,\Omega \] The intended multiplier would have been \[ R_{\mathrm{intended}}=10000-100 \] \[ R_{\mathrm{intended}}=9900\,\Omega \] \(\textbf{Calculate the actual installed resistance:}\) The fitted multiplier is \[ R_{\mathrm{actual}}=9400\,\Omega \] The complete instrument therefore has resistance \[ R_{V,\mathrm{actual}}=G+R_{\mathrm{actual}} \] \[ R_{V,\mathrm{actual}}=100+9400 \] \[ R_{V,\mathrm{actual}}=9500\,\Omega \] \(\textbf{Determine the actual full-scale voltage:}\) Full-scale deflection still requires \(1.0\,\text{mA}\). Hence, \[ V_{\mathrm{FS,actual}}=I_gR_{V,\mathrm{actual}} \] \[ V_{\mathrm{FS,actual}}=(1.0\times10^{-3})(9500) \] \[ V_{\mathrm{FS,actual}}=9.5\,\text{V} \] \(\textbf{Determine the indicated reading at }8.0\,\text{V}\textbf{:}\) The actual current is \[ I=\frac{8.0}{9500} \] \[ I=0.8421\,\text{mA} \] The pointer therefore reaches the fraction \[ \frac{I}{I_g}=\frac{0.8421}{1.0} \] \[ \frac{I}{I_g}=0.8421 \] The incorrectly marked scale assigns \(10\,\text{V}\) to full-scale deflection. Its indicated voltage is \[ V_{\mathrm{ind}}=10\frac{I}{I_g} \] \[ V_{\mathrm{ind}}=10(0.8421) \] \[ V_{\mathrm{ind}}\approx8.42\,\text{V} \] \(\textbf{Check the direction of calibration error:}\) The installed resistance is smaller than intended. A given applied voltage therefore produces greater current and deflection. The marked scale consequently overestimates the actual voltage. \(\textbf{Final answer:}\) \[ \boxed{V_{\mathrm{FS,actual}}=9.5\,\text{V},\qquad V_{\mathrm{ind}}\approx8.42\,\text{V}} \] An incorrect multiplier changes the relation between applied voltage and galvanometer current. It does not change the intrinsic full-scale current of the moving coil. The instrument reaches maximum deflection before the intended \(10\,\text{V}\) value. Its scale error is systematic because the pointer response remains linear but the scale factor is incorrect.

Explanation: \(\textbf{Determine the converted voltmeter resistance:}\) A \(10\,\text{V}\) full-scale reading corresponds to the galvanometer's full-scale current: \[ I_g=2.5\times10^{-3}\,\text{A} \] Therefore, \[ R_V=\frac{10}{I_g} \] \[ R_V=\frac{10}{2.5\times10^{-3}} \] \[ R_V=4000\,\Omega \] The galvanometer already contributes \(100\,\Omega\). Thus, the required series multiplier is \[ R=R_V-G \] \[ R=4000-100 \] \[ R=3900\,\Omega \] \(\textbf{Calculate the undisturbed load voltage:}\) Before connecting the voltmeter, the load and the source's series resistance form a voltage divider. The total resistance is \[ R_{\mathrm{initial}}=2000+4000 \] \[ R_{\mathrm{initial}}=6000\,\Omega \] The current is \[ I_0=\frac{12}{6000} \] \[ I_0=2.0\,\text{mA} \] Therefore, the original load voltage is \[ V_0=I_0(4000) \] \[ V_0=8.0\,\text{V} \] \(\textbf{Determine the load resistance after connecting the voltmeter:}\) The voltmeter is connected in parallel with the \(4.0\,\text{k}\Omega\) load. Its input resistance is also \(4.0\,\text{k}\Omega\). Their equivalent resistance is \[ R_p=\frac{(4000)(4000)}{4000+4000} \] \[ R_p=2000\,\Omega \] \(\textbf{Calculate the voltage in the loaded circuit:}\) The source's series resistance remains \(2000\,\Omega\). The complete circuit resistance becomes \[ R_{\mathrm{loaded}}=2000+2000 \] \[ R_{\mathrm{loaded}}=4000\,\Omega \] The new source current is \[ I=\frac{12}{4000} \] \[ I=3.0\,\text{mA} \] The voltage across the parallel load–voltmeter combination is \[ V=IR_p \] \[ V=(3.0\times10^{-3})(2000) \] \[ V=6.0\,\text{V} \] The voltmeter indicates this new terminal voltage. \(\textbf{Final answer:}\) \[ \boxed{R=3.9\,\text{k}\Omega,\qquad V_0=8.0\,\text{V},\qquad V_{\mathrm{reading}}=6.0\,\text{V}} \] The converted voltmeter has finite input resistance and therefore draws current. Connecting it across the load changes the effective load resistance. The circuit's potential difference is consequently disturbed by the measurement. The voltmeter accurately indicates the voltage present after connection, but that value differs from the undisturbed load voltage. An ideal voltmeter would have infinite resistance and would not produce this loading effect.

Explanation: The current sensitivity of a moving-coil galvanometer is \[ S_I=\frac{\phi}{I} \] For a radial-field galvanometer, this quantity depends on the coil geometry, magnetic field and torsional restoring constant. Adding a resistance externally in series does not directly alter those physical parameters. When the converted voltmeter is connected across an applied voltage \(V\), the current through the moving coil is \[ I=\frac{V}{R_V} \] Therefore, \[ \phi=S_I\frac{V}{R_V} \] Statement I is correct. The voltage sensitivity of the converted instrument is \[ S_V=\frac{\phi}{V}=\frac{S_I}{R_V} \] Increasing the multiplier increases total resistance \(R_V\). Consequently, the deflection produced by a given applied voltage decreases. At the same time, the full-scale voltage is \[ V_g=I_gR_V \] It increases because the galvanometer's full-scale current remains unchanged. Statement II is correct. The intrinsic current sensitivity does not contain the multiplier resistance. Adding the multiplier changes current drawn from an applied voltage, not angular deflection per unit current through the moving coil. Statement III is incorrect. Different multipliers can provide different voltage ranges using the same moving-coil mechanism. For each range, full-scale deflection still occurs when the current through that coil reaches \(I_g\). Statement IV is correct. It is important to distinguish current sensitivity, voltage sensitivity and maximum voltage range. Increasing range by adding resistance generally sacrifices deflection per volt. Greater measuring range is not equivalent to greater sensitivity.

Explanation: \(\textbf{Determine the resistance of the parallel measuring arrangement:}\) The resistor being investigated has resistance \[ R=6.0\,\text{k}\Omega \] The voltmeter has finite input resistance \[ R_V=12\,\text{k}\Omega \] Because they are connected in parallel, \[ R_p=\frac{RR_V}{R+R_V} \] Substituting, \[ R_p=\frac{(6000)(12000)}{6000+12000} \] \[ R_p=4000\,\Omega \] \(\textbf{Determine the total source current:}\) The ammeter is connected in series with this parallel combination. Its effective resistance is \[ R_A=2000\,\Omega \] The total circuit resistance is \[ R_{\mathrm{total}}=R_A+R_p \] \[ R_{\mathrm{total}}=2000+4000 \] \[ R_{\mathrm{total}}=6000\,\Omega \] Therefore, the ammeter indicates \[ I_A=\frac{18}{6000} \] \[ I_A=3.0\,\text{mA} \] \(\textbf{Determine the voltmeter reading:}\) The voltmeter measures the voltage across the parallel combination. Therefore, \[ V=I_AR_p \] \[ V=(3.0\times10^{-3})(4000) \] \[ V=12\,\text{V} \] \(\textbf{Determine the current through the unknown resistor:}\) The resistor current is \[ I_R=\frac{V}{R} \] \[ I_R=\frac{12}{6000} \] \[ I_R=2.0\,\text{mA} \] The remaining current flows through the voltmeter: \[ I_V=\frac{12}{12000} \] \[ I_V=1.0\,\text{mA} \] Thus, \[ I_R+I_V=3.0\,\text{mA} \] which matches the ammeter reading. \(\textbf{Identify the student's incorrect inference:}\) The student calculates \[ R_{\mathrm{inferred}}=\frac{V}{I_A} \] \[ R_{\mathrm{inferred}}=\frac{12}{3.0\times10^{-3}} \] \[ R_{\mathrm{inferred}}=4.0\,\text{k}\Omega \] This is the resistance of the parallel combination, not the unknown resistor alone. \(\textbf{Final answer:}\) \[ \boxed{V=12\,\text{V},\qquad I_A=3.0\,\text{mA},\qquad R_{\mathrm{inferred}}=4.0\,\text{k}\Omega} \] The ammeter measures current through the resistor and voltmeter together. The finite voltmeter resistance allows an additional branch current. Consequently, dividing the voltmeter reading by the total ammeter reading gives the equivalent parallel resistance. The actual resistor current must be distinguished from the total circuit current. Correct interpretation depends on knowing precisely where each measuring instrument is connected.

Explanation: \(\textbf{Determine the required series multiplier:}\) The full-scale current is \[ I_g=1.0\times10^{-3}\,\text{A} \] The target voltmeter range is \[ V_g=20\,\text{V} \] At full scale, \[ V_g=I_g(G+R) \] Therefore, \[ R=\frac{20}{1.0\times10^{-3}}-200 \] \[ R=20000-200 \] \[ R=19800\,\Omega \] \(\textbf{Calculate current at }15\,\text{V}\textbf{:}\) The complete voltmeter resistance is \[ R_V=20000\,\Omega \] Therefore, \[ I_{15}=\frac{15}{20000} \] \[ I_{15}=0.75\times10^{-3}\,\text{A} \] \(\textbf{Determine multiplier heating at the operating voltage:}\) The multiplier dissipates power \[ P_R=I^2R \] Substituting, \[ P_{R,15}=(0.75\times10^{-3})^2(19800) \] \[ P_{R,15}=0.0111375\,\text{W} \] \[ P_{R,15}\approx11.14\,\text{mW} \] \(\textbf{Determine heating at full-scale voltage:}\) At full scale, the moving-coil current is \[ I_{\mathrm{FS}}=1.0\times10^{-3}\,\text{A} \] The multiplier carries the same current because it is in series. Thus, \[ P_{R,\mathrm{FS}}=I_{\mathrm{FS}}^2R \] \[ P_{R,\mathrm{FS}}=(1.0\times10^{-3})^2(19800) \] \[ P_{R,\mathrm{FS}}=0.0198\,\text{W} \] \[ P_{R,\mathrm{FS}}=19.8\,\text{mW} \] \(\textbf{Check the physical consequence:}\) The multiplier dissipates almost all the voltmeter's electrical power because it is much more resistive than the galvanometer coil. Its power rating must exceed the expected operating dissipation. A resistor rated below \(19.8\,\text{mW}\) would be unsuitable for continuous full-scale operation under these idealised conditions. \(\textbf{Final answer:}\) \[ \boxed{R=19.8\,\text{k}\Omega,\qquad P_{R,15}\approx11.14\,\text{mW},\qquad P_{R,\mathrm{FS}}=19.8\,\text{mW}} \] The multiplier protects the galvanometer by limiting current but must itself dissipate electrical power. Power varies as the square of current in a fixed resistor. Therefore, operating at three-quarters of full-scale voltage produces only nine-sixteenths of the full-scale multiplier power. The resistor's electrical resistance and its permissible power rating are separate design requirements.

Explanation: \(\textbf{Determine the voltage required across the moving coil:}\) The galvanometer has \[ G=100\,\Omega \] Its full-scale coil current is \[ I_g=1.0\,\text{mA} \] Therefore, its full-scale terminal voltage is \[ V_g=I_gG \] \[ V_g=(1.0\times10^{-3})(100) \] \[ V_g=0.100\,\text{V} \] \(\textbf{Determine the current through the retained shunt:}\) The shunt is connected across the same terminals. Its resistance is \[ S=100\,\Omega \] At the full-scale galvanometer voltage, \[ I_s=\frac{V_g}{S} \] \[ I_s=\frac{0.100}{100} \] \[ I_s=1.0\,\text{mA} \] \(\textbf{Calculate the current through the multiplier:}\) The multiplier is connected in series with the complete parallel combination. Therefore, it carries the sum of the two branch currents: \[ I_R=I_g+I_s \] \[ I_R=1.0+1.0 \] \[ I_R=2.0\,\text{mA} \] \(\textbf{Determine the voltage available across the multiplier:}\) The desired full-scale voltage of the complete voltmeter is \[ V_{\mathrm{FS}}=20\,\text{V} \] The galvanometer–shunt combination already has \(0.100\,\text{V}\) across it. Therefore, \[ V_R=20.0-0.100 \] \[ V_R=19.9\,\text{V} \] \(\textbf{Calculate the multiplier resistance:}\) \[ R=\frac{V_R}{I_R} \] \[ R=\frac{19.9}{2.0\times10^{-3}} \] \[ R=9950\,\Omega \] \[ R=9.95\,\text{k}\Omega \] \(\textbf{Verify using equivalent resistance:}\) The retained parallel combination has resistance \[ R_p=\frac{(100)(100)}{100+100} \] \[ R_p=50\,\Omega \] The complete voltmeter resistance is \[ R_V=R+R_p \] \[ R_V=9950+50 \] \[ R_V=10000\,\Omega \] At \(20\,\text{V}\), it draws \[ I=\frac{20}{10000}=2.0\,\text{mA} \] \(\textbf{Final answer:}\) \[ \boxed{R=9.95\,\text{k}\Omega,\qquad I_{\mathrm{total,FS}}=2.0\,\text{mA}} \] The retained shunt divides the current entering the sensitive parallel combination. Consequently, the complete voltmeter draws twice the galvanometer's coil current at full scale. The standard bare-galvanometer multiplier formula cannot be applied without accounting for the shunt. Leaving the shunt attached also produces a lower input resistance than an ordinary \(20\,\text{V}\) voltmeter made directly from the same galvanometer.

Explanation: A practical voltmeter draws a small current from the circuit across which it is connected. The current drawn depends on the actual terminal voltage and the meter's input resistance. For an applied voltage \(V\), \[ I_V=\frac{V}{R_V} \] The resistance of voltmeter P is one-fifth that of Q. Therefore, at the same actual terminal voltage, P draws five times the current drawn by Q. Statement I is correct. Connecting a voltmeter across a circuit element creates an additional parallel branch. That branch can change the equivalent resistance seen by the source. When the source has nonzero series resistance, the altered current distribution can change the voltage being measured. A higher-resistance voltmeter generally produces less loading. Statement II is correct. Equal full-scale voltage ratings describe the maximum calibrated measurement range. They do not establish equality of instrument resistance or current drawn. Two voltmeters can share the same voltage range while having substantially different loading effects. Statement III is incorrect. At an actual terminal voltage of \(8.0\,\text{V}\), P draws \[ I_P=\frac{8.0}{10000}=0.80\,\text{mA} \] while Q draws \[ I_Q=\frac{8.0}{50000}=0.16\,\text{mA} \] Their currents are therefore unequal even though their indicated voltages are equal. Statement IV is incorrect. The current drawn by a voltmeter must not be inferred from its full-scale voltage alone. Both range and input resistance are required. For most voltage measurements, a larger input resistance is desirable because it reduces disturbance of the original circuit.

Explanation: \(\textbf{Determine the resistance of the converted voltmeter:}\) The voltmeter has full-scale voltage \[ V_g=10\,\text{V} \] and galvanometer full-scale current \[ I_g=1.0\,\text{mA} \] Its complete input resistance is \[ R_V=\frac{V_g}{I_g} \] \[ R_V=\frac{10}{1.0\times10^{-3}} \] \[ R_V=10000\,\Omega \] \(\textbf{Account for the incorrect series connection:}\) The external resistor has resistance \[ R=100\,\Omega \] Because the voltmeter has been connected in series with it, \[ R_{\mathrm{total}}=R_V+R \] \[ R_{\mathrm{total}}=10000+100 \] \[ R_{\mathrm{total}}=10100\,\Omega \] \(\textbf{Calculate the circuit current:}\) The ideal battery maintains \[ V=12\,\text{V} \] Therefore, \[ I=\frac{12}{10100} \] \[ I\approx1.188\times10^{-3}\,\text{A} \] \[ I\approx1.19\,\text{mA} \] \(\textbf{Determine the external resistor's voltage:}\) \[ V_R=IR \] \[ V_R=\frac{12}{10100}(100) \] \[ V_R\approx0.1188\,\text{V} \] \(\textbf{Interpret the voltmeter's actual terminal voltage:}\) The potential difference across the converted voltmeter is \[ V_{\mathrm{meter}}=IR_V \] \[ V_{\mathrm{meter}}=\frac{12}{10100}(10000) \] \[ V_{\mathrm{meter}}\approx11.88\,\text{V} \] This exceeds its nominal \(10\,\text{V}\) measuring range. The actual pointer behaviour beyond full scale is not specified. The ideal resistance calculation nevertheless shows that almost the entire battery voltage appears across the incorrectly connected voltmeter. \(\textbf{Final answer:}\) \[ \boxed{I\approx1.19\,\text{mA},\qquad V_R\approx0.119\,\text{V},\qquad V_{\mathrm{meter}}\approx11.88\,\text{V}} \] A voltmeter must be connected in parallel with the element whose potential difference is to be measured. Connecting it in series introduces its large input resistance into the main circuit. This severely reduces the current and changes the circuit operation. The resulting potential difference across the voltmeter is not the potential difference across the external resistor. The incorrect connection may also drive the moving coil beyond its calibrated range.

Explanation: A converted galvanometer voltmeter consists of the moving coil and a large resistance connected in series. The instrument is placed across the potential difference to be measured. Under normal operation, the total series resistance limits the current through the sensitive moving coil. The full-scale range is determined by the voltage required to drive the galvanometer's permissible full-scale current through this complete resistance. Under Fault P, the multiplier branch becomes open. Because the elements are in series, this breaks the entire current path through the moving coil. An applied potential difference can still exist between the instrument's external terminals, but the steady current through the galvanometer becomes zero. The pointer can therefore remain at zero. Under Fault Q, the multiplier is bypassed by a short circuit. The effective instrument resistance becomes approximately the galvanometer resistance alone. At the same applied terminal voltage, a much larger current may now pass through the moving coil. This can exceed the galvanometer's full-scale current and may damage it. The Assertion is therefore true. The Reason correctly identifies the purpose of the multiplier and the different electrical consequences of opening and shorting it. It also directly explains both parts of the Assertion. The two faults cannot be treated as equivalent simply because both involve failure of the same resistor. An open multiplier suppresses current, whereas a shorted multiplier can produce dangerously excessive current. This contrast is opposite in an important respect to faults in the shunt branch of a converted ammeter.

Explanation: \(\textbf{Determine the input resistance of each range:}\) The galvanometer full-scale current is \[ I_g=1.0\,\text{mA} \] For a converted voltmeter, \[ R_V=\frac{V_{\mathrm{FS}}}{I_g} \] On the \(5\,\text{V}\) range, \[ R_{V,5}=\frac{5.0}{1.0\times10^{-3}} \] \[ R_{V,5}=5000\,\Omega \] On the \(20\,\text{V}\) range, \[ R_{V,20}=\frac{20}{1.0\times10^{-3}} \] \[ R_{V,20}=20000\,\Omega \] \(\textbf{Calculate the reading on the }5\,\text{V}\textbf{ range:}\) The load resistance is \[ R_L=10000\,\Omega \] The voltmeter is connected in parallel with it. The parallel equivalent is \[ R_{p,5}=\frac{R_LR_{V,5}}{R_L+R_{V,5}} \] \[ R_{p,5}=\frac{(10000)(5000)}{10000+5000} \] \[ R_{p,5}=\frac{10000}{3}\,\Omega \] The source's series resistor is \(5000\,\Omega\). The voltage across the parallel combination is \[ V_5=9.0\frac{R_{p,5}}{5000+R_{p,5}} \] \[ V_5=9.0\frac{10000/3}{5000+10000/3} \] \[ V_5=3.6\,\text{V} \] \(\textbf{Calculate the reading on the }20\,\text{V}\textbf{ range:}\) The voltmeter resistance is now \(20000\,\Omega\). The parallel equivalent becomes \[ R_{p,20}=\frac{(10000)(20000)}{10000+20000} \] \[ R_{p,20}=\frac{20000}{3}\,\Omega \] The new load voltage is \[ V_{20}=9.0\frac{20000/3}{5000+20000/3} \] \[ V_{20}=9.0\frac{20000}{35000} \] \[ V_{20}=\frac{36}{7}\,\text{V} \] \[ V_{20}\approx5.14\,\text{V} \] \(\textbf{Compare the disturbance caused by the two ranges:}\) Without any voltmeter, the load voltage would be \[ V_0=9.0\frac{10000}{5000+10000} \] \[ V_0=6.0\,\text{V} \] The \(20\,\text{V}\) range gives the reading closer to this undisturbed value because its input resistance is larger. Both readings are within their selected calibrated ranges. \(\textbf{Final answer:}\) \[ \boxed{V_5=3.6\,\text{V},\qquad V_{20}=\frac{36}{7}\,\text{V}\approx5.14\,\text{V}} \] Changing a voltmeter's range can change its input resistance. The resulting change in loading can alter the potential difference present across the circuit element. Thus, the two ranges need not indicate identical voltages when used separately on a circuit with appreciable source resistance. The higher range causes less disturbance here, even though it gives less pointer deflection per volt.

Explanation: A galvanometer measures small currents through its sensitive moving coil. To convert it into an ammeter, a small shunt resistance is connected in parallel with the coil. Most of the measured branch current can then bypass the moving coil. The effective resistance of the converted instrument is reduced because of the parallel combination. The ammeter is connected in series with the branch whose current is to be measured. Therefore, P matches property 2. To convert a galvanometer into a voltmeter, a large resistance is connected in series with the coil. This limits the current drawn from the measured circuit for a specified applied potential difference. The converted instrument is connected in parallel across the circuit element or pair of points of interest. Its input resistance is comparatively large. Therefore, Q matches property 1. An ideal ammeter should not alter the circuit current when inserted in series. Its resistance is therefore zero, so it produces no insertion voltage drop. Thus, R matches property 3. An ideal voltmeter should not draw current from the circuit across which it is connected. Its input resistance is therefore infinite. Thus, S matches property 4. The two instrument conversions deliberately use opposite resistance arrangements. The ammeter obtains a small effective resistance by parallel shunting. The voltmeter obtains a large effective resistance by series multiplication. Both are designed to minimise disturbance of the quantity being measured, but they do so in different ways. Confusing their connections can significantly alter circuit behaviour and may damage the instrument.

Explanation: \(\textbf{Current-sensitivity change:}\) The current sensitivity of a radial-field moving-coil galvanometer is \[ S_I=\frac{NAB}{C} \] The number of turns, coil area and magnetic field remain unchanged. Only the torsional constant is doubled: \[ C'=2C \] Therefore, \[ S'_I=\frac{NAB}{2C} \] \[ S'_I=\frac12S_I \] The intrinsic current sensitivity is halved. \(\textbf{Voltmeter sensitivity under the same change:}\) The converted voltmeter has total input resistance \[ R_V=G+R \] \[ R_V=100+9900 \] \[ R_V=10000\,\Omega \] This resistance does not change during the spring replacement. The voltage sensitivity is \[ S_V=\frac{S_I}{R_V} \] Since \(S_I\) halves and \(R_V\) remains constant, \[ S'_V=\frac12S_V \] Thus, voltage sensitivity is also halved. \(\textbf{Full-scale range:}\) Now consider the full-scale range. The full-scale angle \(\phi_g\) is unchanged. Because \[ S_I=\frac{\phi_g}{I_g} \] halving sensitivity requires twice the current to reach the same maximum angle: \[ I'_g=2I_g \] The full-scale voltage of the converted voltmeter is \[ V_g=I_gR_V \] After the spring replacement, \[ V'_g=I'_gR_V \] \[ V'_g=2I_gR_V \] \[ V'_g=2V_g \] Therefore, the full-scale voltage range doubles. A stiffer spring makes the instrument less sensitive because more current is required for the same angular deflection. That same reduction in sensitivity permits measurement of a greater voltage before full-scale deflection. The external series multiplier has not changed, so the increased range arises entirely from the altered mechanical restoring torque.

Explanation: \(\textbf{Determine the required multiplier for P:}\) The target full-scale voltage is \[ V_g=10\,\text{V} \] The required total voltmeter resistance is \[ R_{V,P}=\frac{V_g}{I_{g,P}} \] \[ R_{V,P}=\frac{10}{0.50\times10^{-3}} \] \[ R_{V,P}=20000\,\Omega \] The coil itself contributes \(100\,\Omega\). Thus, \[ R_P=20000-100 \] \[ R_P=19900\,\Omega \] \(\textbf{Determine the required multiplier for Q:}\) Its full-scale current is \[ I_{g,Q}=1.0\,\text{mA} \] The required total resistance is \[ R_{V,Q}=\frac{10}{1.0\times10^{-3}} \] \[ R_{V,Q}=10000\,\Omega \] The moving-coil resistance is \(200\,\Omega\). Therefore, \[ R_Q=10000-200 \] \[ R_Q=9800\,\Omega \] \(\textbf{Compare source currents at the same applied voltage:}\) Both voltmeters are separately connected across \(5.0\,\text{V}\). For P, \[ I_P=\frac{5.0}{20000} \] \[ I_P=0.25\,\text{mA} \] For Q, \[ I_Q=\frac{5.0}{10000} \] \[ I_Q=0.50\,\text{mA} \] Therefore, \[ \frac{I_P}{I_Q}=\frac12 \] \(\textbf{Compare angular deflections:}\) Both instruments have the same full-scale deflection: \[ \phi_g=0.40\,\text{rad} \] P operates at \[ \frac{I_P}{I_{g,P}}=\frac{0.25}{0.50}=\frac12 \] Q operates at \[ \frac{I_Q}{I_{g,Q}}=\frac{0.50}{1.0}=\frac12 \] Therefore, their angular deflections are \[ \phi_P=\phi_Q=\frac12(0.40) \] \[ \phi_P=\phi_Q=0.20\,\text{rad} \] \(\textbf{Final answer:}\) \[ \boxed{R_P=19.9\,\text{k}\Omega,\quad R_Q=9.8\,\text{k}\Omega,\quad I_P=0.25\,\text{mA},\quad I_Q=0.50\,\text{mA},\quad \phi_P=\phi_Q=0.20\,\text{rad}} \] The two voltmeters have identical full-scale voltage ranges and full-scale angular deflections. Consequently, their calibrated pointer fractions agree at the same applied ideal-source voltage. However, their input resistances differ because their galvanometers have different full-scale currents. They therefore draw different currents from the source. Equal indicated voltages and equal pointer deflections do not imply equal loading of the measured circuit.