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Kendriya Vidyalaya Sangathan Ahmedabad Region. Question Bank (Unit II - Current Electricity ) Very Short Answer Type Questions ( 1 Mark ) 1. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature? (Ans : drift velocity decrease) 2. The colours of four bands are yellow, violet, brown and gold. What is the resistance with tolerance limit. ( Ans 470 β¦ ± 5% ) 3. How does the relaxation time of electron in the conductor change when temperature of the conductor decreases ? (ans : T Ξ± 1/π , π increases ) 4. Two wire of equal length one copper and manganin have same resistance , which wire is thicker? (ans: for same R and l , Ο Ξ± A => Ο menganin>ΟCu => A menganin> A Cu =>menganin is thicker ) 5. In the given graph of voltage vs current for a semiconductor, Identify the negative resistance region. (ans : BC region) Short Answer Type Questions ( 2 Marks ) 1. Establish a relation between current and drift velocity. Let n = no density of free electrons π πππ΄ππ£π Ans : current flowing in a conductor πΌ = π‘ = ππ πΌ = πππ΄π£π π π£π 2. Define resistivity of a conductor. Draw the variation of resistivity versus temperature for (i) Nichrome (ii) Silicon Ans : Resistivity is equal to resistance of a conductor having unit length and unit cross sectional area. 3. What happens to the drift velocity (vd) of electrons and to the resistance R if length of a conductor is doubled (keeping potential difference unchanged)? Justify. ππ π¬ ππ π½ π ( Ans : vd halved since vd = π = π π ππππππππ ππ β π and R becomes four π times since πΉ = π π¨ = π ππ π¨ π = 4 times of initial value) 4. Draw a plot showing the variation of terminal voltage (V) vs the current (I) drawn from the cell. Using this plot, how does one determine the internal resistance of the cell ? Ans: π = πΈ β πΌπ π€βπππ πΈ ππ πππ ππ π‘βπ ππππ. πΌππ‘πππππ πππ ππ π‘ππππ πππ ππ πππππ’πππ‘ππ ππ¦ πππππππ π ππππ ππ π‘βπ ππππ. 5. Find the value of the unknown resistance X and the current drawn by the circuit from the battery if no current flows through the galvanometer. Assume the resistance per unit length of the wire is 0.01Ξ© cm-1. π 120 Ans: = ππ π = 3 β¦ 2 80 Short Answer Type Questions ( 3 Marks ) 1. State Kirchhoffβs Rules for an electrical circuit. Hence obtain the balanced condition of Wheatstone bridge. Ans: (i) Junction Rule: Amount of current leaving a junction is equal to amount of current entering the junction. (ii) Loop Rule: algebric sum of change in potential around a closed loop of an electric circuit is zero. Wheastone Bridge: ( P, Q, R and S are four resistors connected in Wheatstone Bridge ) Applying Loop rule in ABCA I1P + IgG - (I-I1) R = 0 β¦β¦β¦β¦β¦..(1) Loop BCDB (I1 β Ig) Q β ( I β I1-Ig ) S β IgG = 0 β¦β¦β¦β¦(2) When the wheatstone bridge is balanced, galavanometer shows no deflection i e Ig = 0 . Therefore equation (1) and (2) becomes I1P = (I-I1) R β¦β¦β¦β¦.(3) & I1Q = ( I β I1) S β¦β¦β¦β¦. (4) Dividing equation (3) by (4) π π = π π 2. (i) Calculate Equivalent Resistance of the given electrical network between points A and B. (ii) Also calculate the current through CD & ACB if a 10V d.c source is connected between point A and B and the value of R = 2ο. Ans: (i) The circuit is a balanced wheatsone bridge. Therefore resistance of CD arm is ineffective. Equivalent Ckt diagram is 1 Effective resistance between A and B is π = π 1 4 1 + 4 ππ π π = 2 β¦ (ii) Current through branch CD is zero as the points C and D are at same potential. 10 Current through ACB branch is πΌπ΄πΆπ΅ = (2+2) = 2.5 π΄ 3. Derive an expression for the resistivity of a good conductor, in terms of the relaxation time of electrons. Solution: Relation between the resistivity and relaxation time: We know that drift velocity of electron is given by: However, β΄ According to Ohmβs law, But resistivity is given by: Comparing (i) and (ii), we obtain The required relationship between resistivity and relaxation time of electrons is . 4. Derive an expression for equivalent emf and internal resistance of two cells connected in parallel. 5. (i) A wire of resistance 4R is bend in the form of circle .What is the effective resistance between the ends of diameter?. (ii) The resistance of a platinum wire at 00C is 4 Ξ©. What will be the resistance at 1000C, if the temperature coefficient of resistance is 0.0038 /0C Ans (i) Effective resistance of each of the two parts will be 2R ohm and the two 1 1 resistors will be connected in parallel. Therefore effective resistance is π = 2π + π 1 ππ π π = R β¦ (ii) π π‘ = π 0 ( 1+ β π‘ ) = 4 ( 1 + 0.0038 × 100 ) = 5.52 β¦ 2π Long Answer Type Questions ( 5 Marks ) 1. State the principle of potentiometer. Explain and draw a circuit diagram used to find internal resistance of a primary cells. Why potentiometer is known as ideal voltmeter? How can the sensitivity of a potentiometer be increased? Ans: Potential difference across any portion of the potentiometer wire is directly proportional to the length of the conductor provided wire is of uniform area of cross section and constant current is flowing through the wire. Measurement of internal resistance of a cell using potentiometer: The cell of emf, E (internal resistance r) is connected across a resistance box (R) through key K . When K is open, balance length is obtained at length AN = l E= Ξ¦ l β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. (1) When K is closed: Let V be the terminal potential difference of cell and the balance is obtained at AN = l β΄ V = Ξ¦ l β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦(2) From equations (1) and (2), we get 2 2 1 1 1 2 2 2 2 From (3) and (4), we get We know l1, l2 and R, so we can calculate r. Potentiometer works on the null deflection method and virtually draws no current from the ckt. (hence behaves as infinite resistance voltmeter) To increase sensitivity of potentiometer its potential gradient should be decreases which can be achieved by (i)increasing length of the wire and (ii)decreasing current in the ckt. 2. State the principle of meter bridge Draw a circuit diagram for a meter bridge two determine the unknown resistance of a resistor. why are the connections between the resistors of a meter bridge made of thick copper strips? Find the shift in the balance point of a meter bridge, when the two resistors in the two gaps are interchanged. Solution: Metrebridge works on the principle of Wheatstone bridge. Meter bridge consists of a metre scale, a wire of length 1 m and a galvanometer. Meter scale is attached to the wire and one end of the galvanometer is connected to the metallic strip between the resistors R and S and the other end is connected to a jockey, which can slide over the metallic meter scale. R is the unknown resistance whose value is to be determined. S is a known resistance. The jockey connected is moved on the metre scale such that we get zero deflection at some point. Let this point be l1 distance away from A. If Rcm is the resistance per unit length of the 1 meter wire, then the resistance of one portion of the wire will be Rcm l1 and the resistance of another portion will be Rcm(100 - l1). Using the condition for balanced bridge, we obtain By substituting the value of l1 in the above equation, we can find out the value of R, which is the unknown resistance. The resistivity of a copper wire of very low. Also, the connections are thick, so that the area is quite large and hence the resistance of the wires is almost negligible. The Balance points will get interchanged.