Current Electricity

a) Consider the circuit in the figure. How much energy is absorbed by electrons from the initial state of no current to the state of drift velocity? b) Electrons give up energy at the rate of RI2 per second to the thermal energy. What time scale would one associate with energy in problem a) n = no of electron/volume = 1029/m3, length of circuit = 10 cm, cross-section = A = 1mm2

a) Current is given as I = V/R from the Ohm’s law Therefore, I = 1A But, I = ne Avd vd = I/neA When the values for the above parameters are substituted, vd = 1/1.6 × 10-4 m/s The KE = (KE of one...

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 In an experiment with a potentiometer, VB = 10V. R is adjusted to be 50 Ω. A student wanting to measure voltage E1 of a battery finds no null point possible. He then diminishes R to 10 Ω and is able to locate the null point on the last segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

Equivalent resistance of the potentiometer = 50 Ohm + R’ Equivalent voltage across the potentiometer = 10 V Current through the main circuit I = 10/(50 Ohms +R’) Potential difference across wire of...

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A room has AC run for 5 hours a day at a voltage of 220V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions?

Power consumption in a day = 10 units Power consumption per hour = 2 units Power consumption = 2 units = 2 kW = 2000 J/s Power consumption in resistors, P = VI Which gives I = 9A We know that...

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Two cells of voltage 10V and 2V and internal resistances 10Ω and 5Ω respectively are connected in parallel with the positive end of the 10V battery connected to the negative pole of 2V battery. Find the effective voltage and effective resistance of the combination.

Kirchhoff’s law is applied at c, I1 = I + I2 Kirchhoff’s law is applied at efbae, 10 – IR – 10I2 = 0 10 = IR + 10I1 Kirchhoff’s law is applied at cbadc, -2-IR+5I2 = 0 2 = 5I2- RI I2 = I1 – I...

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Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistance RA to RB.

Resistance of wire R = ρ l/A Where A is the cross-sectional area of the conductor L is the length of the conductor ρ is the specific resistance RA = ρl/π(10-3 × 0.5)2 RB = ρl/ π[10-3)2 × (0.5 ×...

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The circuit in the figure shows two cells connected in opposition to each other. Cell E1 is of emf 6V and internal resistance 2Ω; the cell E2 is of emf 4V and internal resistance 8 Ω. Find the potential difference between the points A and B.

Applying Ohm’s law, equivalent emf of the two cells = 6 – 4 = 2V Equivalent resistance = 2 + 8 = 10 Ω Electric current, I = 6-4/2+8 = 0.2A When the loop is considered in the anti-clockwise...

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Two cells of same emf E but internal resistance r1 and r2 are connected in series to an external resistor R. What should be the value of R so that the potential difference across the terminals of the first cell becomes zero.

Effective emf of two cells = E + E = 2E Effective resistance = R + r1 + r2 Electric current is given as I = 2E/R+r1+r2 Potential difference is given as V1 – E – Ir1 = 0 Which f=gives R = r1 –...

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Let there be n resistors R1……..Rn with Rmax = max(R1……Rn) and Rmin = min(R1…….Rn). Show that when they are connected in parallel, the resultant resistance Rp < Rmin and when they are connected in series, the resultant resistance Rs > Rmax. Interpret the result physically.

  The current is represented as I = E/R+nR when the resistors are connected in series. Current is expressed as 10I = E/(R+R/n) when the resistors are connected in parallel....

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First, a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘n’?

The current is represented as I = E/R+nR when the resistors are connected in series. Current is expressed as 10I = E/(R+R/n) when the resistors are connected in parallel. We get n = 10 by solving...

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While doing an experiment with potentiometer it was found that the deflection is one-sided and i) the deflection decreased while moving from one end A of the wire to the end B; ii) the deflection increased, while the jockey was moved towards the end B. i) Which terminal +ve or –ve of the cell E, is connected at X in case
i) and how is E1 related to E?
ii) Which terminal of the cell E1 is connected at X in case ii)?

The positive terminal of cell E1 is linked to E, and E is connected to X.  Furthermore, E1 > E ii) cell E1's negative terminal is linked to X.

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The relaxation time τ is nearly independent of applied E field whereas it changes significantly with temperature T. First fact is responsible for Ohm’s law whereas the second fact leads to a variation of ρ with temperature. Elaborate why?

Relaxation time is the time interval between two successive collisions of the electrons.It is defined asτ = mean free path/rms velocity of electrons usually, the drift velocity of the electrons is...

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Temperature dependence of resistivity ρ(T) of semiconductors, insulators, and metals is significantly based on the following factors:
a) number of charge carriers can change with temperature T
b) time interval between two successive collisions can depend on T
c) length of material can be a function of T
d) mass of carriers is a function of T

The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T

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In a meter bridge, the point D is a neutral point.
a) the meter bridge can have no other neutral point for this set of resistances
b) when the jockey contacts a point on meter wire left of D, current flows to B from the wire
c) when the jockey contacts a point on a meter wire to the right of D, current flows from B to the wire through the galvanometer
d) when R is increased, the neutral point shifts to left

The correct answer is a) the meter bridge can have no other neutral point for this set of resistances c) when the jockey contacts a point on a meter wire to the right of D, current flows from B to...

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Temperature dependence of resistivity ρ(T) of semiconductors, insulators, and metals is significantly based on the following factors:
a) number of charge carriers can change with temperature T
b) time interval between two successive collisions can depend on T
c) length of material can be a function of T
d) mass of carriers is a function of T

solution:The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T

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Consider a simple circuit in the figure.stands for a variable resistance R’.
R’ can vary from R0 to infinity. r is internal resistance of the battery,
a) potential drop across AB is nearly constant as R’ is varied
b) current through R’ is nearly a constant as R’ is varied
c) current I depends sensitively on R’
d) I ≥V/r+R always

solution: The correct answer is a) potential drop across AB is nearly constant as R’ is varied d) I ≥V/r+R always  

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Kirchhoff’s junction rule is a reflection of
a) conservation of current density vector
b) conservation of charge
c) the fact that the momentum with which a charged particle approaches a junction is unchanged as the charged particle leaves the junction
d) the fact that there is no accumulation of charges at a junction

solution: The correct answer is b) conservation of charge d) the fact that there is no accumulation of charges at a junction

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A metal rod of length 10 cm and a rectangular cross-section of 1 cm × 1/2 cm is connected to battery across opposite faces. The resistance will be
a) maximum when the battery is connected across 1 cm × 1/2 cm faces
b) maximum when the battery is connected across 10 cm × 1 cm faces
c) maximum when the battery is connected across 10 cm × 1/2 cm faces
d) same irrespective of the three faces

solution:The correct solution is a) maximum when the battery is connected across 1 cm × 1/2 cm faces

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Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400 cm.
a) the battery that runs the potentiometer should have voltage of 8V
b) the battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V
c) the first portion of 50 cm of wire itself should have a potential drop of 10V
d) potentiometer is usually used for comparing resistances and not voltages

Solution: The correct solution is b) the potentiometer's battery can be set to 15V and R adjusted so that the potential drop across the wire is a little higher than 10V.

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A resistance R is to be measured using a meter bridge. Student chooses the standard resistance S to be 100Ω. He finds the null point at l1 = 2.9 cm . He is told to attempt to improve the accuracy. Which of the following is a useful way?
a) he should measure l1 more accurately
b) he should change S to 1000 Ω and repeat the experiment
c) he should change S to 3 Ω and repeat the experiment
d) he should give up hope of a more accurate measurement with a meter bridge

solution:The correct answer is c) he should change S to 3 Ω and repeat the experiment

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Two batteries of emf ε1 and ε2 and internal resistances r1 and r2 respectively are connected in parallel as shown in the figure.a) the equivalent emf εeq of the two cells is between ε1 and ε2 that is ε1 < εeq < ε2
b) the equivalent emf εeq is smaller than ε1
c) the εeq is given by εeq = ε1 + ε2 always
d) εeq is independent of internal resistances r1 and r2

      solution: The correct answer is a) the equivalent emf εeq of the two cells is between ε1 and ε2 that is ε1 < εeq < ε2

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Consider a current-carrying wire in the shape of a circle. Note that as the current progresses along the wire, the direction of j changes in an exact manner, while the current I remain unaffected. The agent that is essentially responsible for is
a) source of emf
b) electric field produced by charges accumulated on the surface of wire
c) the charges just behind a given segment of wire which push them just the right way by repulsion
d) the charges ahead

solution: The correct answer is b) electric field produced by charges accumulated on the surface of wire

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Figure below shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

Answer – According to the question statement – Internal resistance of the cell is r = 1.5 V cell Balance point of the cell in open circuit is l = 76.3 cm External resistance is R = 9.5 Ω...

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Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value ε?  (b) What purpose does the high resistance of 600 kΩ have? (c) Is the balance point affected by this high resistance  (d) Would the method work in the above...

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(a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in figure

Answer – (a) The total number of resistors is equal to n. Each resistor's resistance is equal to R.  (i) When the resistors are connected in series, the maximum effective resistance is...

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Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (1022/1023).

Solution: (a) greater (b) lower (c) nearly independent of (d) 1022

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Answer the following questions:
(a) A steady current flows in a metallic conductor of the non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements that do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

Answer: (a) The current is assumed to be constant. As a result, it's a constant. The area of cross-section has an inverse relationship with current density, electric field, and drift speed. (b) No,...

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Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (ρAl = 2.63 × 10–8 Ω m, ρCu = 1.72 × 10–8 Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Answer – According to the question statement, some given properties of aliminium are – Length = l1 Resistance = R Resistivity ,ρAI =ρ1= 2.63×10−8 Ωm Relative density , d1 = 2.7 Area...

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(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What is the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Answer – According to the question statement, (a) Emf of the secondary cells is ε = 2.0 V N is the number of secondary cells  = 6 Then total EMF is given by – E = nε = 6 x 2 E = 12 V r  =...

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The earth’s surface has a negative surface charge density of 10–9 C m–2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire
globe. If there were no mechanism of sustaining atmospheric electric field, how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe). (Radius of earth = 6.37 × 10m.)

Answer – According to the question statement, Surface charge density of the earth is σ = 10−9 cm−2Potential difference between the surface and the top of the atmosphere is V= 400 kVCurrent over...

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The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 1028 m–3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10–6 m2 and it is carrying a current of 3.0 A.

Answer – It is given that Number density of free electrons in a copper conductor is n =  8.5 x 10 28 m – 3 Assume that the Length of the copper wire is denoted by l and we have l...

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A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the
purpose of having a series resistor in the charging circuit?

Answer – According to the question statement – The EMF of storage battery is E = 8.0 V Internal resistance of battery is given by r = 0.5 Ω DC supply voltage is V = 120 V Resistance of the resistor...

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A ) In a meter bridge given below, the balance point is found to be at 39.5 cm from the end A, when the resistor S is of 12.5 Ω. Determine the resistance of R. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

B ) Determine the balance point of the bridge above if R and S are interchanged. C ) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the...

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A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70 × 10
–4 °C–1.

Answer – According to the question; Supply voltage,  V = 230 V initial current drawn is given by I 1 = 3.2 A Let the initial resistance be given by R 1, which can be determined...

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At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1.

Answer – We are given that,  Room temperature, T = 27 ° C Resistance of heating element , R = 100 Ω Let the increased temperature of the filament be given by T 1 At T 1, the...

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