Current Electricity (Previous Year Questions)

Q1.  heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few second to a steady value of 2.8 A. what is the steady temperature of the heating element if the room temperature is 27.0 ^oC ? temperature coefficient of nichrome average over the temperature range involved is 1.70 x 10^{-4 o}C^-1.

Ans. Here V = 230 V and at $T_1 = 27$ degree celcius, current   $ I_1 = 3.2 $A \[R_1 = \frac{V}{I_1} =\frac{230}{3.2} \Omega\]

Again at a steady temperature $ T_2$ of the heating element, current $I_2 = 2.8 $ A \[R_2 = \frac{V}{I_2} =\frac{230}{2.8} \Omega\]

Moreover temperature coefficient of resistance $ \alpha = (1.70 * {10^{-4}}) ^{\circ} C^{-1}$

Using the relation  \[R_2 = R_1 [1 + \alpha (T_2 -T_1)]\], we have \[T_2 -T_1 = \frac{R_2 - R_1}{R_1 \alpha} = 840\] \[T_2 = T_1 + 840 = 27 +840 = 867^{\circ}C\]

Q2. (a)    In a metre bridge, the balance point is found to be at 39.5 cm from the end A containing X toward end A, when the resistor Y is of 12.5$\Omega$. Determine the resistance of X. 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 X and Y are interchanged.

(c)    What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? would the galvanometer shown any current?

Ans. (a) here Y = 12.5 ohm, length AD = $l_1 = 39.5$ cm \[\frac{X}{Y} = \frac{l_1}{100-l_1}\] \[X = Y\frac{l_1}{100-l_1} = 12.5*\frac{39.5}{60.5} = 8.2 \Omega\]

Connection are made of thick copper strips so that their resistance may be extremely small and negligible, because these resistances are not accounted for in the formula of meter bridge.

(b) let on interchanging X and Y, the new balance point is obtained at $l_2$, then\[\frac{Y}{X} = \frac{l_2}{100-l_2} \implies l_2 = 60.5 cm\]

(c) At the balance point at the bridge if the galvanometer at the cell are interchanged, it makes no effect on balance condition and the galvanometer will not show any deflection.

Q3. State the condition in which terminal voltage across a secondary cell is equal to its emf.

Ans. When the cell is in an open circuit i.e., when no current is being drawn from the cell.

Q4. Under what condition can we draw maximum current from a secondary cell?

Ans. When external resistance present in the circuit is zero i.e., when the cell is short circuited.

Q5. A wire of resistivity $\rho$ is stretched to twice its length. What will be its new resistivity?

Ans. Resistivity will remain unchanged, because resistivity of a material is independent of its dimensions.

Q6. A physical quantity, associated with electric conductivity, has the SI unit ‘’ohm-meter.”  Identify the physical quantity.

Ans. Resistivity.

Q7. Define electrical conductivity of a conductivity of the conductor and give its SI unit.

Ans. Reciprocal of resistivity of a conductor is called its conductivity. Alternatively conductance of a unit cube conductor is called its electric conductivity. Its SI unit is S m-1.

Q8. If potential difference V applied across a conductor is increased to 2 V , how will the drift velocity of the electrons change ?

Ans. Drift speed   \[v_d = \frac{eE}{m}\tau = \frac{eV}{ml}\tau\]Thus, it is clear that on increasing the potential difference from V to 2V, the drift speed of the electrons is doubled.

Q9. What is the effect of heating of a conductor of a drift velocity of a free electrons?

Ans. On heating a conductor its resistance increase or the current decreases. Consequently, the drift velocity of free electron decreases.

Q10. If the temperature of a good conductor increase, how does the relaxation time of electrons in the conductor change?

Ans. With increase in temperature the resistivity of conductor material increases and hence in accordance with the formula $ \rho = \frac{m}{ne^2\tau} $, the relaxation period $\tau$ decreases.

Q11. 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.

Ans. It is given that number density of electrons in X is twice that in Y, i.e., $ n_x = 2n_y $. As in a series circuit the electric current flowing through the entire circuit is exactly same, Hence \[I =n_xA_X e(v_d)_X = n_YA_Ye (v_d)_y\]As both wire have same diameter, hence $ A_x = A_y $ \[\frac{(v_d)_x}{(v_d)_y} =\frac{n_y}{n_x}= \frac{n_y}{2n_y} = 0.5\]

Q12. Two wires of equal length, one of copper and other of manganin have the same resistance. Which wire is thicker ?

Ans. In accordance with the formula $ R = \rho \frac{L}{A} $ for same resistance R and length l, \[A \propto \rho\]. Hence, the manganin wire will be thicker because its resistivity is more.

Q13. Write an expression for the resistivity of a metallic conductor showing its variation over a limited range of temperatures.

Ans. $\rho_T = \rho_0[ 1 + \alpha(T - T_0) ] $, where $\alpha$ is the temperature coefficient of resistivity.

Q14. Why are alloys, maganin and constantan used to make standard resistance coils ?

Ans. Because their resistivity is high and temperature coefficient of resistance is extremely small.

Q15. The metallic conductor is at a temperature $\theta_1$. The temperature of the metallic conductor is increased to $\theta_2$. How will the product of its resistivity and conductivity change ?

Ans. The product of resistivity and resistivity and conductivity always remains constant

Q16. The three coloured bands on a carbon resistor are red, green and yellow respectively. Write the value of its resistance.

Ans. Value of given resistance is $25*10^4 + 20%  \Omega $.

Q17. The sequence of bands marked on a carbon resistor are : Brown, black, green and gold. Write the value of resistance with tolerance.

Ans. Resistance R = $ 10^6 $ ohm $\pm$ 5%.

Q18. Which physical quantity does the voltage vs. current graph for a metallic conductor depict ? Give its SI unit

Ans. Electrical resistance is given by the slope of V – I graph. Its SI unit is a ohm.

Q19. A(i) series, (ii) parallel combination of two given resistor is connected, one by one, across a cell. In which case will the terminal potential difference, across the cell, have a higher value?

Ans. Terminal potential difference V = E – Ir, where r is the internal resistance of the cell. If two given resistor be $R_1$ and $R_2$ than in series $I_s = \frac{e}{(R_1 + R_2 + r)}$ but in parallel combination current $ I_p = \frac{e}{(\frac{ R_1R_2}{R_1+R_2})+ r }$. Obiviously, $I_s < I_p $. Hence Vs >Vp.

Q20. A cell of emf 2 V and internal resistance $0.1\Omega$ is connected to a $3.9\Omega $ external resistance. What will be the potential difference across the terminals of the cell?

Ans. Terminal potential difference \[V = \frac{eR}{R+r} =\frac{2*3.9}{3.9+0.1}= 1.95 V\]

Q21.What happens to the power dissipation if the value of electric current passing through  conductor of constant resistance in doubled?

Ans. In accordance with formula $ P = I^2R $, the dissipation becomes 4 times if the current passing through a given resistance is doubled.

Q22. Which has a greater resistance, 1 kW electric heater or a 100 W electric bulb, both marked for 200V? 

Ans. Electric bulb marked 220 V – 100W will have higher resistance because its power is less and power is given by $ P = \frac{V^2}{R} \implies R = \frac{V^2}{P} $.

Q23. Two bulb whose resistance are in the ratio of 1:2 are connected in parallel to a square of constant voltage. What will be the power dissipation in these?

Ans. Here V = constant and $\frac{R_1}{R_2} = \frac{1}{2} $, hence  $ \frac{P1}{P2} = \frac{V^2/R_1}{V^2/ R_2} = \frac{R_2}{R_1} = 2$.

Q24. A toaster produces more heat than a light bulb when connected in parallel to the 220 v mains. Which of the two has greater resistance ?

Ans. From the relation $ P = \frac{V^2}{R}$, it is clear that the resistance of bulb is greater as it produces less heat (i.e., its power is less) for constant potential difference.

Q25. Two bulbs are marked 60 W, 220 V, and 100 W, 220 V. These are connected in parallel to 220 V mains. Which one out of the two will glow brighter ?

Ans. Bulb marked 100W, 220V will glow brighter because its power is more.

Q26. Two conductors one having resistance R and another 2R are connected in turn across a d. c. source . If the rate of heat produced in the two conductors is $Q_1$ and $Q_2$ respectively, what is the value of $\frac{Q_1}{Q_2}$ ?

Ans. Here V = const., hence, $ \frac{Q_1}{Q_2} = \frac{R_2}{R_1} = \frac{2R}{R} = 2:1 $.

Q27. A heater joined in series with a 60 W bulb is connected to the mains. If 60 W bulb is replaced by a 100 W bulb, will the rate of heat produced by the heater be more, less or remain the same ?

Ans. We know that resistance of a 100W bulb is less than that of 60 W bulb. Hence, on joining 100 W bulb (instead of 60 W bulb) with heater, the resistance of the circuit decreases and consequently, circuit current increases. Hence, heat produced by the heater rises.

Q28. Two heater wires of the same dimensions are first connected in series and then in parallel to a source of supply. What will be the ratio of heat produced in the two cases ?

Ans. Let resistance of each heater be R then in series arrangement $R_S = 2R $ and in parallel arrangement $R_P = \frac{R}{2} $. In accordance with formula $ H = \frac{V^2t}{R} $, ratio of heat produced in two cases: \[\frac{H_{series}}{H_{parallel}} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4}\]

Q29. Establish a relation between current and drift velocity.

Ans. Consider a conductor of uniform cross-section area A, carrying a current I. Consider a small section KL of the conductor having a length $\Delta x$ or having a volume $ A.\Delta x$, then number of free electrons present in this section = $n A\Delta x$, where n = Number density of free electrons.

Total charge carried by these electrons while crossing the given section $\Delta Q = nAe\Delta x$

Now total time taken by the electrons to cross this section is $ \Delta t = \frac{\Delta x}{v_d} $Where $v_d$ = drift velocity of electrons

By definition \[I=\frac{\Delta Q}{\Delta t} = \frac{nAe\Delta x}{\Delta t} = neAv_d\]

Q30. Derive an expression for the current density of a conductor in terms of the drift speed of electrons.

Or

Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons.

Ans.  Current density \[J = \frac{I}{A} = \frac{neAv_d}{A} = nev_d\] Thus $ J \propto v_d $.

Q31. Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.

Ans. We know that in the absence of an external electric field E, the conduction electrons in a conductor move randomly with velocities $ u_1, u_2, u_3, ….u_n$ such that their mean value \[\frac{u_1 +u_2+u_3+.....+u_n}{n} =0\]

However, in the presence of an external field E, electrons experience an acceleration \[\vec{a} = -\frac{e \vec{E}}{m}\]If $ t_1, t_2, t_3,…. $ be the times before two successive collisions for different electrons, then the final velocities acquired by different electrons are\[\vec{v_1} = \vec{u_1}+\vec{a}t_1, \vec{v_2} = \vec{u_2}+\vec{a}t_1, ...... \vec{v_n} = \vec{u_n}+\vec{a}t_n\]

                            

Mean value of electron velocity in the presence of an electrical field = Drift velocity $\vec{v_d} $ 

\[\frac{\vec{v_1}+\vec{v_2}+....+\vec{v_n}}{n} = \frac{\vec{u_1}+\vec{u_2}+...+\vec{u_n}}{n} + \vec{a}(\frac{t_1+t_2+.....+t_n}{n})\]

\[\vec{v_d} = \vec{a}\tau = -\frac{e\vec{E}}{m}\tau\]

Where relaxation time.\[\tau = \frac{t_1+t_2+...+t_n}{n}\]