CBSE Class 12 Physics Current Electricity Assignment Set 04

UNIT 2

CURRENT ELECTRICITY

 Important Questions for NCERT Class 12 Physics Current Electricity

Question. An electric bulb is rated 60 W, 220 V. The resistance of its filament is   
(a) 870 W
(b) 780 W
(c) 708 W
(d) 807 W 

Answer: D

Question. A current of 2 A, passing through a conductor produces 80 J of heat in 10 seconds. The resistance of the conductor in ohm is    
(a) 0.5
(b) 2
(c) 4
(d) 20

Answer: B

Question. A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is    
(a) 230 W
(b) 46 W
(c) 26 W
(d) 13 W

Answer: C

Question. A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be    
(a) 2 A
(b) 1 A
(c) 0.5 A
(d) 0.25 A

Answer: C

Question. Faraday law of electrolysis indirectly shows
(a) quantisation of charge 
(b) quantisation of angular momentum
(c) quantisation of current
(d) quantisation of viscosity

Answer: A

Question. Two sources of equal emf are connected to an external resistance R. The internal resistance of the two sources are R1 and R2 (R2 > R1). If the potential difference across the source having internal resistance R2 is zero, then 
(a) R = R₂ – R₁
(b) R = R2 × (R₁ + R₂)/(R₂ – R₁)
(c) R = R₁R₂/(R₂ – R₁)
(d) R = R₁R₂/(R₁ – R₂)

Answer: A

Question. When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be   
(a) R/4 
(b) 4R
(c) R/8 
(d) R/2 

Answer: A

Question. Resistance n, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to   
(a) n2R
(b) R/n2
(c) R/n
(d) nR 

Answer: A

Question. When three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be   
(a) 60 watt
(b) 180 watt
(c) 10 watt
(d) 20 watt 

Answer: D

Question. Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt of power.    
What will be the power dissipated in watt if the same resistors are connected in parallel across the same source of e.m.f.?
(a) 30
(b) 10 /3
(c) 10
(d) 90

Answer: D

Question. One kilowatt hour is equal to   
(a) 36 × 10^–5 J
(b) 36 × 10^–4 J
(c) 36 × 10⁵ J
(d) 36 × 10³ J 

Answer: C

Question. A 4 mF capacitor is charged to 400 V. If its plates are joined through a resistance of 2 kW, then heat produced in the resistance is   
(a) 0.64 J
(b) 1.28 J
(c) 0.16 J
(d) 0.32 J 

Answer: D

Question. If two bulbs, whose resistances are in the ratio of 1 : 2 are connected in series, the power dissipated in them has the ratio of   
(a) 2 : 1
(b) 1 : 4
(c) 1 : 1
(d) 1 : 2.

Answer: D

Question. 50 electric bulbs are connected in series across a 220 V supply and the illumination produced is I₁. 5 bulbs are fused. If the remaining 45 are again connected in series, the illumination produced is I₂. Which of the following is true ?
(a) I₁ = I₂
(b) I₁ < I₂
(c) I₁ > I₂
(d) It will depend on the resistance of each bulb.

Answer: B

Question. A leclanche cell supplies a current of one ampere for ten minutes. The electreochemical equivalent of hydrogen = 0.00001014 gram per coulomb. The mass of hydrogen liberated is :
(a) 0.00625 g
(b) 0.01248 g
(c) 0.01872 g
(d) 0.02496 g.

Answer: A

Question. Three equal resistors connected across a source of e.m.f. together dissipate 10 watt of power. What will be the power dissipated in watt if the same resistors are connected in parallel across the same source of e.m.f.?
(a) 10/3
(b) 10
(c) 30
(d) 90

Answer: D

Question. In an experiment to measure the internal resistance of a cell, by a potentiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5 Ω resistance and is at a length of 3 m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is then
(a) 1.5 Ω
(b) 10 Ω
(c) 15 Ω
(d) 1 Ω

Answer: B

Question. Twelve resistors each of resistance 16 Ω are connected in the circuit as shown. The net resistance between AB is
(a) 1Ω
(b) 2 Ω
(c) 3 Ω
(d) 4 Ω

Answer: D

Question. A 5–ampere fuse wire can withstand a maximum power of 1 watt in the circuit. The resistance of the fuse wire is 
(a) 0.04 Ω
(b) 0.2 Ω
(c) 5 Ω
(d) 0.4 Ω

Answer: A

Question. The cold junction of a thermocouple is maintained at 10ºC. No thermo e.m.f. is developed when the hot junction is maintained at 530ºC. The neutral temperature is
(a) 260ºC
(b) 265ºC
(c) 270ºC
(d) 520ºC

Answer: C

Question. Three equal resistors connected across a source of e.m.f. together dissipate 10 watt of power. What will be the power dissipated in watts if the same resistors are connected in parallel across the same source of e.m.f.?
(a) 10 
(b) 10/3
(c) 30
(d) 90

Answer: D

Question. An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be
(a) 15 min
(b) 8 min
(c) 4 min
(d) 25 min

Answer: B

Question. 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%

Answer: C

Question. A wire of resistance 4 Ω is stretched to twice its original length. The resistance of stretched wire would be
(a) 4 Ω
(b) 8 Ω
(c) 16 Ω
(d) 2 Ω

Answer: C

Question. The charge flowing through a resistance R varies with time t as Q = at – bt², where a and b are positive constants. The total heat produced in R is:
(a) a³B/6b
(b) a³B/3b
(c) a³B/2b
(d) a³B/b

Answer: A

Question. There is an infinite wire grid with cells in the form of equilateral triangles. The resistance of each wire between neighbouring joint connections is R₀. The net resistance of the whole grid between the points A and B as shown is 

(a) R₀
(b) R₀/2
(c) R₀/3
(d) R₀/4

Answer: C

Question. Two 1000 W heaters when connected in parallel across 220 V supply produced heat QP in time t. If they are connected in series across the same power supply the heat produced in the same time is QS. What is QP/QS?
(a) 4
(b) 2
(c) 0.5
(d) 0.25.

Answer: A

Question. Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be 

(a) 2V/R
(b) 3V/R
(c) V/R
(d) V/2R

Answer: A

Question. A battery is charged at a potential of 15V for 8 hours when the current flowing is 10A. The battery on discharge supplies a current of 5A for 15 hour. The mean terminal voltage during discharge is 14V. The “Watt-hour” efficiency of the battery is
(a) 87.5%
(b) 82.5%
(c) 80%
(d) 90%

Answer: A

Question. When three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be
(a) 20 watt
(b) 60 watt
(c) 180 watt
(d) 10 watt

Answer: A

Question. In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be
(a) R/2
(b) R
(c) 2R
(d) R/4

Answer: D

Question. In the network shown, each resistance is equal to R. The equivalent resistance between adjacent corners A and D is 
(a) R 
(b) 2/3 R
(c) 3/7 R
(d) 8/15 R

Answer: D

Question. A battery of e.m.f E and internal resistance r is connected to a variable resistor R as shown. Which one of the following is true ?
(a) Potential difference across the terminals of the battery is maximum where R = r
(b) Power delivered to resistor is maximum when R = r
(c) Current in the circuit is maximum when R = r
(d) Current in the circuit is maximum when R >> r

Answer: B,C

Question. A potentiometer wire of length L and a resistance r are connected in series with a battery of e.m.f. E₀ and a resistance r₁. An unknown e.m.f. E is balanced at a length l of the potentiometer wire. The e.m.f. E will be given by: 

Answer: A

Question. In the circuit shown in figure, the 5W resistance develops 20.00 cal/s due to the current flowing through it. The heat developed in 2 Ω resistance (in cal/s) is
(a) 23.8
(b) 14.2
(c) 11.9
(d) 7.1

Answer: B

Question. If specific resistance of a potentiometer wire is 10–7Ωm current flowing through it, is 0.1 amp and cross sectional area of wire is 10–6 m2, then potential gradient will be
(a) 10^–2 volt/m
(b) 10^–4 volt/m
(c) 10^–6 volt/m
(d) 10^–8 volt/m

Answer: A

Question. Cell having an emf e and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by 

Answer: C

Question. A milli voltmeter of 25 milli volt range is to be converted into an ammeter of 25 ampere range. The value (in ohm) of necessary shunt will be
(a) 0.001
(b) 0.01
(c) 1
(d) 0.05

Answer: A

Question. In the circuit shown the cells A and B have negligible resistances. For V_A = 12V, R₁ = 500Ω and R = 100Ω the galvanometer (G) shows no deflection. The value of V_B is
(a) 4 V
(b) 2 V
(c) 12 V
(d) 6 V

Answer: B

Question. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf's is : 
(a) 5 : 1
(b) 5 : 4
(c) 3 : 4
(d) 3 : 2

Answer: D

Question. A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves 
(a) Potential gradients
(b) A condition of no current flow through the galvanometer
(c) A combination of cells, galvanometer and resistances
(d) Cells

Answer: B

Question. The resistance of the four arms P, Q, R and S in a Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90 ohm, respectively. The e.m.f. and internal resistance of the cell are 7 volt and 5 ohm respectively. If the galvanometer resistance is 50 ohm, the current drawn from the cell will be
(a) 0.2 A
(b) 0.1 A
(c) 2. 0 A
(d) 1. 0 A

Answer: A

Question. A heating coil is labelled 100 Ω, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source.
The energy now liberated per second is
(a) 25 J
(b) 50 J
(c) 200 J
(d) 400 J

Answer: D

Question. The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be :-}
(a) R/n
(b) n²R
(c) R/n²
(d) nR

Answer: B

Question. If 25W, 220 V and 100 W, 220 V bulbs are connected in series across a 440 V line, then
(a) only 25W bulb will fuse
(b) only 100W bulb will fuse
(c) both bulbs will fuse
(d) None of these

Answer: A

Question. In the circuit shown in the figure, if potential at point A is taken to be zero, the potential at point B is
(a) –1V
(b) + 2V
(c) –2V
(d) + 1V

Answer: D

Question 1. The electron drift speed in a metallic conductor is only a few mm \(\text{s}^{-1}\) for currents in the range of the few amperes. How then is the current established almost at the instant the circuit is closed?
Answer: The electric field is established throughout the conductor at the speed of electromagnetic waves (which is nearly \(3 \times 10^8 \text{ m/s}\)).
This electric field exerts a force on every free electron everywhere in the conductor almost instantaneously.
Therefore, drift of electrons starts at all points in the circuit almost at the same time, establishing the current instantly.
In simple words: The electrical signal travels at the speed of light, making all free electrons start moving simultaneously.

Exam Tip: Mention that the speed of signal propagation is much greater than the drift speed of electrons.

Question 2. When electrons drift in a metal from lower to higher potential, does it mean that the free electrons of the metal are moving in the same direction?
Answer: No.
Free electrons move in random thermal motion constantly colliding with ions.
The net directional drift of electrons is opposite to the electric field, which is from lower potential to higher potential.
However, individual electrons have random thermal velocities in all directions, and only experience a small net drift velocity opposite to the electric field.
In simple words: Individual electrons move randomly in all directions, but their overall average drift is towards the higher potential.

Exam Tip: Distinguish clearly between random thermal motion and net directional drift.

Question 3. Two conducting wires X and Y of same diameter but different materials are joined in series across a battery. If the number density of electrons in X is twice that in Y, find the ratio of drift velocity of electrons in the two wires.
Answer: Current \(I\) through both wires in series is the same: \(I_X = I_Y = I\).
The current is given by \(I = n e A v_d\), where \(n\) is number density, \(e\) is electronic charge, \(A\) is cross-sectional area, and \(v_d\) is drift velocity.
Since \(I\), \(e\), and \(A\) are the same for both wires:
\(n_X v_{dX} = n_Y v_{dY}\)
Given that \(n_X = 2 n_Y\):
\(2 n_Y \cdot v_{dX} = n_Y \cdot v_{dY}\)
\(\frac{v_{dX}}{v_{dY}} = \frac{1}{2}\)
In simple words: Since current is constant and wire X has twice the electron density, its electrons must drift at half the speed of wire Y.

Exam Tip: Use the principle of conservation of current in a series circuit: \(I_1 = I_2\).

Question 4. Two wires one of manganin and the other of copper have equal length and equal resistance. Which one of these wires will be thicker?
Answer: Resistance is given by \(R = \rho \frac{l}{A}\), where \(\rho\) is resistivity, \(l\) is length, and \(A\) is cross-sectional area.
Area \(A = \frac{\pi d^2}{4}\), so \(R = \frac{4 \rho l}{\pi d^2}\).
For equal length (\(l\)) and equal resistance (\(R\)):
\(d^2 \propto \rho\)
Since the resistivity of manganin (\(\rho_m\)) is much greater than the resistivity of copper (\(\rho_c\)), i.e., \(\rho_m > \rho_c\):
The diameter of the manganin wire is greater than that of the copper wire.
Therefore, the manganin wire will be thicker.
In simple words: Manganin has higher resistivity, so to have the same resistance as copper for the same length, it needs a larger cross-sectional area (be thicker).

Exam Tip: Remember that resistivity (\(\rho\)) of alloys like manganin is much higher than that of pure metals like copper.

Question 5. Two \(120\text{ V}\) light bulbs, one of \(25\text{ W}\) and the other of \(200\text{ W}\) were connected in series across a \(240\text{ V}\) line. One bulb burnt out almost instantaneously. Which one was burnt and why?
Answer: Resistance of a bulb is given by \(R = \frac{V^2}{P}\).
For the \(25\text{ W}\) bulb: \(R_{25} = \frac{120^2}{25} = \frac{14400}{25} = 576 \, \Omega\).
For the \(200\text{ W}\) bulb: \(R_{200} = \frac{120^2}{200} = \frac{14400}{200} = 72 \, \Omega\).
When connected in series across \(240\text{ V}\), the current \(I\) is the same through both.
Voltage across the \(25\text{ W}\) bulb is \(V_{25} = I \times 576\), and across the \(200\text{ W}\) bulb is \(V_{200} = I \times 72\).
Since \(R_{25} > R_{200}\), a much larger voltage drops across the \(25\text{ W}\) bulb, exceeding its rated voltage of \(120\text{ V}\), causing it to burn out instantaneously.
In simple words: The \(25\text{ W}\) bulb has much higher resistance, so it takes a larger share of the total voltage and fuses instantly.

Exam Tip: In series combination, higher resistance draws higher potential drop (\(V = IR\)).

Question 6. A cell of emf \(E\) and internal resistance \(r\) is connected across an external resistance \(R\). Plot a graph showing the variation of P.D. across \(R\), verses \(R\).
Answer: The potential difference \(V\) across external resistance \(R\) is given by:
\(V = IR = \frac{E}{R + r} R = \frac{E}{1 + \frac{r}{R}}\)
When \(R = 0\), \(V = 0\).
When \(R \to \infty\), \(V \to E\).
Graph description: The curve starts from the origin \((0,0)\) and asymptotically approaches the value \(E\) as \(R\) increases towards infinity.
In simple words: As external resistance increases, the voltage across it increases and gradually approaches the total EMF of the cell.

Exam Tip: Ensure the graph shows an asymptotic saturation line at \(V = E\).

Question 7. Draw the graphs showing the variation of resistivity with temperature for (i) nichrome and (ii) silicon.
Answer: (i) For nichrome (alloy): Resistivity increases almost linearly and very slowly with an increase in temperature.
(ii) For silicon (semiconductor): Resistivity decreases exponentially with an increase in temperature due to an increase in charge carrier density.
In simple words: Alloys show a slight steady increase in resistance with temperature, whereas semiconductors show a sharp decrease.

Exam Tip: Clearly label axes (\(\rho\) vs \(T\)) and curve trends for metal alloys versus semiconductors.

Question 8. The circuit shown in the diagram contains a battery \(\text{B}\), a rheostat \(\text{Rh}\) and identical lamps \(\text{P}\) and \(\text{Q}\). What will happen to the brightness of the lamps, if the resistance through the rheostat is increased? Give reasons.
Answer: Lamp \(\text{P}\) is in series with the main circuit, while lamp \(\text{Q}\) is in parallel with the rheostat branch.
When the resistance of the rheostat (\(\text{Rh}\)) is increased, the total equivalent resistance of the circuit increases.
Consequently, the total current drawn from the battery decreases.
Since lamp \(\text{P}\) carries the total main circuit current, its brightness decreases.
The potential difference across the parallel combination of lamp \(\text{Q}\) and rheostat increases because the voltage drop across the rest of the circuit decreases, but detailed analysis shows that the current through lamp \(\text{Q}\) actually increases (or its voltage increases), hence the brightness of lamp \(\text{Q}\) increases.
In simple words: Increasing rheostat resistance reduces main current (dimming P) but redirects more current/voltage into Q (brightening Q).

Exam Tip: Analyze series and parallel branches separately by tracking current and potential changes.

Question 9. Determine the current drawn from a \(12\text{ V}\) supply with internal resistance \(0.5 \, \Omega\) by the infinite network shown in fig. Each resistor has \(1 \, \Omega\) resistance.
Answer: Let the equivalent resistance of the infinite network be \(R_{\text{eq}}\).
Since the network is infinite, adding one more repeating unit to the front does not change the total equivalent resistance \(R_{\text{eq}}\).
Looking at the ladder network, each repeating section has two \(1 \, \Omega\) resistors in series on the upper and lower arms, and one \(1 \, \Omega\) resistor in parallel.
Thus, \(R_{\text{eq}} = 1 + 1 + \frac{1 \times R_{\text{eq}}}{1 + R_{\text{eq}}} = 2 + \frac{R_{\text{eq}}}{1 + R_{\text{eq}}}\)
\(R_{\text{eq}} - 2 = \frac{R_{\text{eq}}}{1 + R_{\text{eq}}}\)
\((R_{\text{eq}} - 2)(1 + R_{\text{eq}}) = R_{\text{eq}}\)
\(R_{\text{eq}} + R_{\text{eq}}^2 - 2 - 2R_{\text{eq}} = R_{\text{eq}}\)
\(R_{\text{eq}}^2 - 2R_{\text{eq}} - 2 = 0\)
Solving for \(R_{\text{eq}}\) using the quadratic formula:
\(R_{\text{eq}} = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 + \sqrt{3} \approx 1 + 1.732 = 2.732 \, \Omega\)
Total resistance of the circuit including internal resistance \(r = 0.5 \, \Omega\):
\(R_{\text{total}} = R_{\text{eq}} + r = 2.732 + 0.5 = 3.232 \, \Omega\)
Current drawn from the supply:
\(I = \frac{E}{R_{\text{total}}} = \frac{12}{3.232} \approx 3.71\text{ A}\)
In simple words: Treat the infinite repeating part as \(R_{\text{eq}}\), solve the quadratic equation, add internal resistance, and apply Ohm's law.

Exam Tip: Infinite ladder networks are solved by setting up a recursive algebraic equation where the tail equals \(R_{\text{eq}}\).

Question 10. In a meter bridge, the null point is found at a distance of \(40\text{ cm}\) from A. If a resistance of \(12 \, \Omega\) is connected in parallel with S, the null point occurs at \(50.0\text{ cm}\) from A. Determine the values of R and S.
Answer: In the first case, null point is at \(l_1 = 40\text{ cm}\).
Using the meter bridge balance condition:
\(\frac{R}{S} = \frac{l_1}{100 - l_1} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}\)
\(S = \frac{3}{2}R \quad \text{--- (Equation 1)}\)
In the second case, a resistance of \(12 \, \Omega\) is connected in parallel with \(S\).
The new equivalent resistance of the right gap is \(S' = \frac{12S}{12 + S}\).
The new null point is at \(l_2 = 50\text{ cm}\):
\(\frac{R}{S'} = \frac{50}{100 - 50} = \frac{50}{50} = 1\)
\(R = S' = \frac{12S}{12 + S} \quad \text{--- (Equation 2)}\)
Substitute \(S = \frac{3}{2}R\) into Equation 2:
\(R = \frac{12 \left(\frac{3}{2}R\right)}{12 + \frac{3}{2}R}\)
\(12 + \frac{3}{2}R = 18\)
\(\frac{3}{2}R = 6\)
\(R = 6 \times \frac{2}{3} = 4 \, \Omega\)
Now find \(S\):
\(S = \frac{3}{2}(4) = 6 \, \Omega\)
In simple words: Set up the standard meter bridge formula for both conditions, substitute the parallel combination, and solve the simultaneous equations for R and S.

Exam Tip: For parallel combinations in bridge arms, compute the equivalent resistance \(S'\) first before applying the balance ratio.