NEET Physics Current Electricity MCQs Set 18

Physics Objective Questions and Answers: Current Electricity

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Access the complete set of multiple-choice questions for Current Electricity below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official NEET textbooks.

These 25 NEET Physics multiple choice questions on Current Electricity come with answers and explanations to help you check your method. Topics include Ohm's law, resistivity, combinations of resistors, Kirchhoff's laws, cells, the potentiometer and the Wheatstone bridge.

Question 1: The current flowing through a wire depends on time as \(I=3{t}^{2}+2t+5.\) The charge flowing through the cross-section of the wire in time from \(t=0\)to \(t=2 \mathrm{sec}\). is
(a) 22 C
(b) 20 C
(c) 18 C
(d) 5 C
Show Answer & Explanation

Answer: (a) 22 C

Explanation:
1. \(I=\frac{dq}{dt}=3{t}^{2}+2t+5\)
2. \(∴dq=(3{t}^{2}+2t+5)dt\)
3. \(∴q=\int_{t=0}^{t=2} (3{t}^{2}+2t+5)dt\)
4. \(=\frac{3{t}^{3}}{3}+\frac{2{t}^{2}}{2}+5t\left|\begin{matrix}2 \\ 0\end{matrix}={t}^{3}+{t}^{2}+5t\right|\begin{matrix}2 \\ 0\end{matrix}=22 C\)

Question 2: The ratio of voltage sensitivity (\({V}_{S}\)) and current sensitivity (\({I}_{s}\)) of a moving coil galvanometer is
(a) \(\frac{1}{G}\)
(b) \(\frac{1}{{G}^{2}}\)
(c) G
(d) \({G}^{2}\)
Show Answer & Explanation

Answer: (a) \(\frac{1}{G}\)

Explanation:
1. Voltage sensitivity \(=\frac{Q}{V}\)
2. Current sensitivity \(=\frac{Q}{I}\)
3. Also, potential difference
4. \(V=IG\)
5. Hence, \(\frac{{V}_{S}}{{I}_{s}}=\frac{Q/V}{Q/I}=\frac{I}{V}=\frac{I}{IG}\)
6. \(∴\frac{{V}_{s}}{{I}_{s}}=\frac{1}{G}\)

Question 3: How much work is required to carry a \(6 μC\) charge from the negative terminal to the positive terminal of a \(9 V\) battery
(a) \(54×{10}^{-3}J\)
(b) \(54×{10}^{-6}J\)
(c) \(54×{10}^{-9}J\)
(d) \(54×{10}^{-12}J\)
Show Answer & Explanation

Answer: (b) \(54×{10}^{-6}J\)

Explanation:
1. \(W=qV=6×{10}^{-6}×9=54×{10}^{-6}J\)

Question 4: We have two wires \(A\) and \(B\) of same mass and same material. The diameter of the wire \(A\) is half of that \(B.\) If the resistance of wire \(A\) is \(24 ohm\) then the resistance of wire \(B\) will be
(a) \(12 Ohm\)
(b) \(3.0 Ohm\)
(c) \(1.5 Ohm\)
(d) None of the above
Show Answer & Explanation

Answer: (c) \(1.5 Ohm\)

Explanation:
1. Same mass, same material \(i.e.\) volume is same or \(Al=\) constant
2. Also, \(R=ρ\frac{l}{A}⇒\frac{{R}_{1}}{{R}_{2}}=\frac{{l}_{1}}{{l}_{2}}×\frac{{A}_{2}}{{A}_{1}}={\left(\frac{{A}_{2}}{{A}_{1}}\right)}^{2}={\left(\frac{{d}_{2}}{{d}_{1}}\right)}^{4}\)
3. \(⇒\frac{24}{{R}_{2}}={\left(\frac{d}{d/2}\right)}^{4}=16⇒{R}_{2}=1.5Ω\)

Question 5: A \(50 ohm\) galvanometer gets full scale deflection when a current of \(0.01 A\) passes through the coil. When it is converted to a \(10 A\) ammeter, the shunt resistance is
(a) \(0.01 Ω\)
(b) \(0.05 Ω\)
(c) \(2000 Ω\)
(d) \(5000 Ω\)
Show Answer & Explanation

Answer: (b) \(0.05 Ω\)

Explanation:
1. \({i}_{g}=i\frac{S}{G+S}⇒\frac{0.01}{10}=\frac{S}{50+S}⇒S=\frac{50}{999}=0.05Ω\)

Question 6: A resistor \({R}_{1}\) dissipates power \(P\) when connected to a certain generator. If the resistor \({R}_{2}\) is put in series with \({R}_{1},\) the power dissipated by \({R}_{1}\)
(a) Decreases
(b) Increases
(c) Remains the same
(d) Any of the above depending upon the relative values of \({R}_{1}\) and \({R}_{2}\)
Show Answer & Explanation

Answer: (a) Decreases

Explanation:
1. Equivalent resistance in the second case \(={R}_{1}+{R}_{2}=R\)
2. Now, we know that \(P∝\frac{1}{R}\)
3. Since in the second case the resistance \(\left({R}_{1}+{R}_{2}\right)\) is higher than that in the first case \(\left({R}_{1}\right)\)
4. Therefore power dissipation in the second case will be decreased

Question 7: If for a thermocouple \({T}_{n}\) is the neutral temperature, \({T}_{c}\) is the temperature of the cold junction and \({T}_{i}\) is the temperature of inversion, then
(a) \({T}_{i}=2{T}_{n}-{T}_{c}\)
(b) \({T}_{n}={T}_{i}-2{T}_{c}\)
(c) \({T}_{i}={T}_{n}-{T}_{c}\)
(d) None of these
Show Answer & Explanation

Answer: (a) \({T}_{i}=2{T}_{n}-{T}_{c}\)

Explanation:
1. \({T}_{n}=\frac{{T}_{i}+{T}_{C}}{2}⇒{T}_{i}=2{T}_{n}-{T}_{C}\)

Question 8: A current of \(2 A\) passing through conductor produces \(80 J\) of heat in \(10 seconds.\) The resistance of the conductor is
(a) \(0.5 Ω\)
(b) \(2 Ω\)
(c) \(4 Ω\)
(d) \(20 Ω\)
Show Answer & Explanation

Answer: (b) \(2 Ω\)

Explanation:
1. \(H={i}^{2}Rt⇒R=\frac{H}{{i}^{2}t}=\frac{80}{4×10}=2 Ω\)

Question 9: A thick wire is stretched, so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire?
(a) 2 : 1
(b) 4 : 1
(c) 3 : 1
(d) 1 : 4
Show Answer & Explanation

Answer: (c) 3 : 1

Explanation:
1. Resistance of the wire is given by
2. \(R=ρ\frac{l}{A}=ρ\frac{{l}^{2}}{Al}=\frac{ρ{l}^{2}}{V}\left(∵Al=V\right)\)
3. So,
4. \(R∝{l}^{2}\)(if density remains same)
5. \(or \frac{{R}^{'}}{R}=\frac{{\left(2l\right)}^{2}}{{\left(l\right)}^{2}}=4\)
6. \({R}^{'}=4R\)
7. Hence, change in resistance
8. =4R-R=3R
9. \(\mathrm{Therefore,}\frac{\mathrm{change\ in\ resistance}}{\mathrm{original\ resistance}}=\frac{3R}{1R}=3:1\)

Question 10: A thermo couple develops \(200μ V\) between \(0℃\) and \(100℃.\) If it develops \(64 μ V\) and \(76 μ V\) respectively between \((0℃-32℃)\) and \((32℃-70℃)\) then what will be the thermo \(emf\) it develops between \(70℃\)and \(100℃\)
(a) \(65 μ V\)
(b) \(60 μ V\)
(c) \(55 μ V\)
(d) \(50 μ V\)
Show Answer & Explanation

Answer: (b) \(60 μ V\)

Explanation:
1. By using \({e}_{0}^{100}={e}_{0}^{32}+{e}_{32}^{70}+{e}_{70}^{100}\)
2. \(⇒200=64+76+{e}_{70}^{100}⇒{e}_{70}^{100}=60 μV\)

Question 11: When connected across the terminals of a cell, a voltmeter measures \(5V\) and a connected ammeter measures \(10 A\) of current. A resistance of \(2 ohm\) is connected across the terminals of the cell. The current flowing through this resistance will be
(a) \(2.5 A\)
(b) \(2.0 A\)
(c) \(5.0 A\)
(d) \(7.5 A\)
Show Answer & Explanation

Answer: (b) \(2.0 A\)

Explanation:
1. Emf \(E=5V\), Internal resistance \(r=\frac{5}{10}=0.5Ω\)
2. Current through the resistance \(i=\frac{5}{(2+0.5)}=2A\)

Question 12: An electric heater boils 1 kg of water in a time \({t}_{1}.\) Another heater boils the same amount of water in a time \({t}_{2}.\) When the two heaters are connected in parallel, the time required by them together to boil the same amount of water is
(a) \({t}_{1}+ {t}_{2}\)
(b) \({t}_{1}{t}_{2}\)
(c) \(\frac{{t}_{1}+{t}_{2}}{2}\)
(d) \(\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}\)
Show Answer & Explanation

Answer: (d) \(\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}\)

Explanation:
1. Let \({R}_{1}\mathrm{and}{R}_{2}\) be the resistances of the coils
2. \(H=\frac{{V}^{2}{t}_{1}}{{R}_{1}}\mathrm{and} H=\frac{{V}^{2}{t}_{2}}{{R}_{2}}\)
3. \(⇒\frac{{t}_{1}}{{R}_{1}}=\frac{{t}_{2}}{{R}_{2}}, ie,\frac{{R}_{2}}{{R}_{1}}=\frac{{t}_{2}}{{t}_{1}} …(i)\)
4. Now in parallel
5. \({R}^{'}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}={R}_{1}\)
6. \(∴H=\frac{{V}^{2}t}{{R}^{'}} …(\mathrm{ii)}\)
7. Now, \(\frac{{V}^{2}t}{{R}^{'}}=\frac{{V}^{2}{t}_{1}}{{R}_{1}}\)
8. \(⇒\frac{t×\left({R}_{1}+ {R}_{2}\right)}{{R}_{1}{R}_{2}}=\frac{{t}_{1}}{{R}_{1}} …(\mathrm{iii)}\)
9. On using Eqs. (i) and (iii), we get
10. \(t=\frac{{t}_{1}{t}_{2}}{{t}_{1}+{t}_{2}}\)

Question 13: A \(2 volt\) battery, a \(15 Ω\) resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is \(5Ω,\) then the potential gradient of the potentiometer wire is
(a) \(0.005 V/cm\)
(b) \(0.05 V/cm\)
(c) \(0.02 V/cm\)
(d) \(0.2 V/cm\)
Show Answer & Explanation

Answer: (a) \(0.005 V/cm\)

Explanation:
1. Potential gradient\(=\frac{e}{(R+{R}_{h}+r)}.\frac{R}{L}\)
2. \(=\frac{2}{(15+5+0)}×\frac{5}{1}=0.5\frac{V }{m}=0.005\frac{V}{cm}\)

Question 14: To get the maximum current from a parallel combination of \(n\) identical cells each of internal resistance \(r\) and external resistance \(R,\) when
(a) \(R≫r\)
(b) \(R≪r\)
(c) \(R=r\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(R≪r\)

Explanation:
1. Cells are joined in parallel when internal resistance is higher than external resistance. \([R<2. \(i=\frac{E}{R+\frac{r}{n}}\)

Question 15: Consider the following two statements \(A\) and \(B,\) and identify the correct choice out of given answers
A. Thermo \(e\).m.f. is minimum at neutral temperature of a thermocouple
B. When two junctions made of two different metallic wires are maintained at different temperatures, an electric current is generated in the circuit
(a) \(A\) is false and \(B\) is true
(b) \(A\) is true and \(B\) is false
(c) Both \(A\) and \(B\) are false
(d) Both \(A\) and \(B\) are true
Show Answer & Explanation

Answer: (a) \(A\) is false and \(B\) is true

Explanation:
1. \(A\) is false because at neutral temperature thermo emf is maximum. \(B\) is true

Question 16: The temperature of cold junction of thermocouple is \(0°C.\) If the neutral temperature is \(270°C,\) then the inversion temperature is
(a) \(540°C\)
(b) \(520°C\)
(c) \(640°C\)
(d) \(58°C\)
Show Answer & Explanation

Answer: (a) \(540°C\)

Explanation:
1. Neutral temperature, \({T}_{n}=\frac{{T}_{i}+ {T}_{c}}{2}\)
2. or \({T}_{i}=2{T}_{n}- {T}_{c}=540℃.\)

Question 17: A battery of emf E and internal resistance r is connected to an external resistance R the condition for maximum power transfer is
(a) r<R
(b) r>R
(c) r=1/R
(d) R=R
Show Answer & Explanation

Answer: (d) R=R

Explanation:
1. We know that the current in the circuit
2. \(I=\frac{E}{R+r}\)
3. and power delivered to the resistance R is
4. \(P={I}^{2}R=\frac{{E}^{2}R}{{\left(R+r\right)}^{2}}\)
5. It is maximum when \(\frac{dP}{dR}=0\)
6. \(\frac{dP}{dR}={E}^{2}\left[\frac{{\left(r+R\right)}^{2}-2R\left(r+R\right)}{{\left(r+R\right)}^{4}}\right]=0\)
7. Or \({\left(r+R\right)}^{2}=2R\left(r+R\right)\)
8. Or \(R=r\)

Question 18: The electrolyte used in Lechlanche cell is
(a) Copper sulphate solution
(b) Ammonium chloride solution
(c) Dilute sulphuric acid
(d) Zinc sulphate
Show Answer & Explanation

Answer: (b) Ammonium chloride solution

Explanation:
1. Just for your knowledge remember, voltaic cell uses dil. \({H}_{2}S{O}_{4} ;\) Dry cell uses \(N{H}_{4}Cl+ZnC{l}_{2}\) paste; Daniel cell uses dil. \({H}_{2}S{O}_{4} ;\) Lead Accumulator uses dil. \({H}_{2}S{O}_{4}\) and \(Ni\)-\(Fe\) cell or Alkaline Accumulator uses \(KOH\) solution

Question 19: At neutral temperature, the thermoelectric power \(\left(\frac{dE}{dT}\right)\) has the value
(a) Zero
(b) Maximum but negative
(c) Maximum but positive
(d) Minimum but positive
Show Answer & Explanation

Answer: (a) Zero

Explanation:
1. At neutral temperature,\(\frac{dE}{dT}=0\)

Question 20: An emf of 0.9 V is generated when the temperature difference hot and cold junction of thermocouple is 75 K. Assuming that the thermo emf is directly proportional to the temperature difference, the extent to which the thermo emf will change when the cold junction is heated up by 15 K is
(a) 10%
(b) 20%
(c) 40%
(d) 60%
Show Answer & Explanation

Answer: (b) 20%

Explanation:
1. Here, \({ E}_{1}=0.9V,θ-{θ}_{0}=75E;{E}_{2}=?\mathrm{when}{θ}_{0}^{'}={θ}_{0}+15\)
2. \(∴\)temperature difference \(θ-{θ}_{0}^{'}=θ-({θ}_{0}+15)\)
3. \(=\left(θ-{θ}_{0}\right)-15=75-15=60\) K
4. As \(\frac{{E}_{2}}{{E}_{1}}=\frac{θ-{θ}_{0}^{'}}{θ-{θ}_{0}}=\frac{60}{75}=\frac{4}{5}\)
5. or % decrease in thermo emf
6. \(=\frac{{E}_{1}-{E}_{2}}{{E}_{1}}×100=\left( 1-\frac{{E}_{2}}{{E}_{1}}\right)×100\)
7. \(=\left( 1-\frac{5}{4}\right)×100=20\%\)

Question 21: If voltage across a bulb rated 220 Volt-100 Watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is
(a) 20%
(b) 2.5%
(c) 5%
(d) 10%
Show Answer & Explanation

Answer: (c) 5%

Explanation:
1. Resistance of bulb is constant
2. \(P=\frac{{v}^{2}}{R}⇒\frac{∆P}{P}=\frac{2∆V}{V}+\frac{∆R}{R}\)
3. \(\frac{∆P}{P}=2×2.5+0=5\%\)

Question 22: For comparing the e.m.f.'s of two cells with a potentiometer, a standard cell is used to develop a potential gradient along the wires. Which of the following possibilities would make the experiment unsuccessful
(a) The e.m.f. of the standard cell is larger than the \(E\) e.m.f.'s the two cells
(b) The diameter of the wires is the same and uniform throughout
(c) The number of wires is ten
(d) The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells
Show Answer & Explanation

Answer: (d) The e.m.f. of the standard cell is smaller than the e.m.f.'s of the two cells

Explanation:
1. The emf of the standard cell must be greater than that of experimental cells, otherwise balance point is not obtained

Question 23: For a thermocouple, the neutral temperature is \(270℃\) and the temperature of its cold junction is \(20℃.\) If there is no deflection in the galvanometer, the temperature of the hot junction should be
(a) \(210℃\)
(b) \(540℃\)
(c) \(520℃\)
(d) \(209℃\)
Show Answer & Explanation

Answer: (c) \(520℃\)

Explanation:
1. When there is no deflection, then this temperature is called inversion temperature. It is given by the relation
2. \({θ}_{n}=\frac{{θ}_{i}+{θ}_{c}}{2}\)
3. Where \({θ}_{c}\) is temperature of cold junction \(=20℃\) and neutral temperature \({θ}_{n}=270℃\)
4. \(∴{θ}_{i}=2{θ}_{n}-{θ}_{c}=540-20=520℃\)

Question 24: A copper wire of cross-sectional area 2.0 \(m{m}^{2}\), resistivity =1.7\(×{10}^{-8}Ωm\), carries a current of 1A. The electric field in the copper wire is
(a) \(8.5×{10}^{-5}V{m}^{-1}\)
(b) \(8.5×{10}^{-4}V{m}^{-1}\)
(c) \(8.5×{10}^{-3}V{m}^{-1}\)
(d) \(8.5×{10}^{-2}V{m}^{-1}\)
Show Answer & Explanation

Answer: (c) \(8.5×{10}^{-3}V{m}^{-1}\)

Explanation:
1. \(E=\frac{V}{l}=\frac{iR}{l}=\frac{iρl}{Al}=\frac{iρ}{A}=\frac{1×1.7×{10}^{-8}}{2×{10}^{-6}}\)
2. \(=8.5×{10}^{-3}V{m}^{-1}\)

Question 25: The electron in a hydrogen atom circles around the proton in \(1.5941×{10}^{-18}\)s. The equivalent current due to motion of the electrons is
(a) 127.37 mA
(b) 122.49 mA
(c) 100.37 mA
(d) 94.037 mA
Show Answer & Explanation

Answer: (c) 100.37 mA

Explanation:
1. The equivalent current due to motion of electrons is given by
2. \(I=\frac{e}{t}=\frac{1.6×{10}^{-19}}{1.594×{10}^{-18}}\)
3. \(=1.0037×{10}^{-1}\)
4. \(=100.37×{10}^{-3}A\)
5. \(=100.37mA\)

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