CBSE 12th Standard Physics Subject Current Electricity Chapter Case Study Questions 2021
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CBSE 12th Standard Physics Subject Current Electricity Case Study Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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The flow of charge in a particular direction constitutes the electric current. Current is measured in Ampere. Quantitatively, electric current in a conductor across an area held perpendicular to the direction of flow of charge is defined as the amount of charge is flowing across that area per unit time.
Current density at a point in a conductor is the ratio of the current at that point in the conductor to the area of cross section of the conductor of that point.
The given figure shows a steady current flows in a metallic conductor of non uniform cross section. Current density depends inversely on area, so, here \(J_{1}>J_{2}, \text { as } A_{1}
(i) What is the current flowing through a conductor, if one million electrons are crossing in one millisecond through a cross-section of it ?(a) 2.5 x 10-10 A (b) 1.6 x 10-10 A (c) 7.5 X 10-9 A (d) 8.2 x 10-11 A (ii) SI unit of electric current is
(a) Cs (b) Ns-2 (c) Cs-1 C-1s-1 (iii) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor?
(a) Electric field (b) Drift velocity (c) Current (d) Current density (iv) A constant current I is flowing along the length of a conductor of variable cross-section as shown in the figure. The quantity which does not depend upon the area of cross-section is
(a) electron density (b) current density (c) drift velocity (d) electric field (v) When a current of 40 A flows through a conductor of area 10 m2, then the current density is
(a) 4 A/m2 (b) 1 A/m2 (c) 2 A/m2 (d) 8 A/m2 -
According to Ohm's law, the current flowing through a conductor is directly proportional to the potential difference across the ends of the conductor i.e \(I \propto V \Rightarrow \frac{V}{I}=R\) where R is resistance of the conductor Electrical resistance of a conductor is the obstruction posed by the conductor to the flow of electric current through it. It depends upon length, area of cross-section, nature of material and temperature of the conductor We can write \(R \propto \frac{l}{A} \text { or } R=\rho \frac{l}{A}\) where \(\rho\) is electrical resistivity of the material of the conductor.
(i) Dimensions of electric resistance is\(\text { (a) }\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-2}\right]\) \(\text { (b) }\left[M L^{2} T^{-3} A^{-2}\right]\) \(\text { (c) }\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{-1} \mathrm{~A}\right]\) \(\text { (d) }\left[M^{-1} L^{2} T^{2} A^{-1}\right]\) (ii) If \(1 \mu \mathrm{A}\) current flows through a conductor when potential difference of2 volt is applied across its ends, then the resistance of the conductor is
\(\text { (a) } 2 \times 10^{6} \Omega\) \(\text { (b) } 3 \times 10^{5} \Omega\) \(\text { (c) } 1.5 \times 10^{5} \Omega\) \(\text { (d) } 5 \times 10^{7} \Omega\) (iii) Specific resistance of a wire depends upon
(a) length (b) cross-sectional area (c) mass (d) none of these (iv) The slope of the graph between potential difference and current through a conductor is
(a) a straight line (b) curve (c) first curve then straight line (d) first straight line then curve (v) The resistivity of the material of a wire 1.0 m long, 0.4 mm in diameter and having a resistance of 2.0 ohm is
\(\text { (a) } 1.57 \times 10^{-6} \Omega \mathrm{m}\) \(\text { (b) } 5.25 \times 10^{-7} \Omega \mathrm{m}\) \(\text { (c) } 7.12 \times 10^{-5} \Omega \mathrm{m}\) \(\text { (d) } 2.55 \times 10^{-7} \Omega \mathrm{m}\) -
Emf of a cell is the maximum potential difference between two electrodes of the cell when no current is drawn from the cell. Internal resistance is the resistance offered by the electrolyte of a cell when the electric current flows through it. The internal resistance of a cell depends upon the following factors;
(i) distance between the electrodes
(ii) nature and temperature of the electrolyte
(iii) nature of electrodes
(iv) area of electrodes.
For a freshly prepared cell, the value of internal resistance is generally low and goes on increasing as the cell is put to more and more use. The potential difference between the two electrodes of a cell in a closed circuit is called terminal potential difference and its value is always less than the emf of the cell in a closed circuit. It can be written as V = E - Jr.
(i) The terminal potential difference of two electrodes of a cell is equal to emf of the cell when\(\text { (a) } I \neq 0\) (b) I=0 (c) both (a) and (b) (d) neither (a)nor (b) (ii) A cell of emf E and internal resistance r gives a current of 0.5 A with an external resistance of \(12 \Omega\) and a current of 0.25 A with an external resistance of \(25 \Omega\) .What is the value of internal resistance of the cell?
\(\text { (a) } 5 \Omega\) \(\text { (b) } 1 \Omega\) \(\text { (c) } 7 \Omega\) \(\text { (d) } 3 \Omega\) (iii) Choose the wrong statement.
(a) Potential difference across the terminals of a cell in a closed circuit is always less than its emf. (b) Internal resistance of a cell decrease with the decrease in temperature of the electrolyte. (c) Potential difference versus current graph for a cell is a straight line with a -ve slope (d) Terminal potential difference of the cell when it is being charged is given as V = E + Ir. (iv) An external resistance R is connected to a cell of internal resistance r, the maximum current flows in the external resistance, when
(a) R = r (b) R < r (c) R> r (d) R=l/r (v) IF external resistance connected to a cell has been increased to 5 times, the potential difference across the terminals of the cell increases from 10 V to 30 V. Then, the emf of the cell is
(a) 30 V (b) 60V (c) 50 V (d) 40 V -
A single cell provides a feeble current. In order to get a higher current in a circuit, we often use a combination of cells A combination of cells is called a battery, Cells can be joined in series, parallel or in a mixed way.
Two cells are said to be connected in series when negative terminal of one cell is connected to positive terminal of the other cell and so on. Two cells are said to be connected in parallel if positive terminal of each cell is connected to one point and negative terminal of each cell connected to the other point. In mixed grouping of cells, a certains number of identical cells are joined in series, and all such rows are then connected in parallel with each other.
(i) To draw the maximum current from a combination of cells, how should the cells be grouped?(a) Parallel (b) Series (c) Mixed grouping (d) Depends upon the relative values of internal and external resistances (ii) The total emf of the cells when n identical cells each of emf e are connected in parallel is
\(\text { (a) } n \varepsilon\) \(\text { (b) } n^{2} \varepsilon\) (c) E \(\text { d) } \frac{\varepsilon}{n}\) (iii) 4 cells each of emf 2 V and internal resistance of \(1 \Omega\) are connected in parallel to a load resistor of \(2 \Omega\). Then the current through the load resistor is
(a) 2 A (b) 1.5 A (c) 1 A (d) 0.888 A (iv) If two cells out of n number of cells each of internal resistance 'r' are wrongly connected in series, then total resistance of the cell is
(a) 2nr (b) nr - 4r (c) nr (d) r (v) Two identical non-ideal batteries are connected in parallel. Consider the following statements.
(i). The equivalent emf is smaller than either of the two emfs.
(ii) The equivalent internal resistance is smaller than either of the two internal resistances(a) Both (i) and (ii) are correct. (b) (i) is correct but (ii) is wrong (c) (ii) is correct but (i) is wrong. (d) Both (i) and (ii) are wrong. -
Potentiometer is an apparatus used for measuring the emf of a cell or potential difference between two points in an electrical circuit accurately. It is also used to determine the internal resistance of a primary cell. The potentiometer is based on the principle that, if V is the potential difference across any portion of the wire of length 1 and resistance R, then \(V \propto l \text { or } V=k l\) where k is the potential gradient. Thus, potential difference across any portion of potentiometer wire is directly proportional to length of the wire of that portion. The potentiometer wire must be uniform. The resistance of potentiometer wire should be high.
(i) Which one of the following is true about potentiometer?(a) Its sensitivity is low (b) It measures the emf of a cell very accurately (c) It is based on deflection method (d) None of the above (ii) A current of 1.0 mA is flowing through a potentiometer wire of length 4 cm and of resistance \(4 \Omega\) .The potential gradient of the potentiometer wire is
\(\text { (a) } 10^{-3} \mathrm{Vm}^{-1}\) \(\text { (b) } 10^{-5} \mathrm{Vm}^{-2}\) \(\text { (c) } 2 \times 10^{-3} \mathrm{Vm}^{-1}\) \(\text { (d) } 4 \times 10^{-3} \mathrm{Vm}^{-1}\) (iii) Sensitivity of a potentiometer can be increased by
(a) decreasing potential gradient along the wire (b) increasing potential gradient along the wire (c) decreasing current through the wire (d) increasing current through the wire (iv) A potentiometer is an accurate and versatile device to make electrical measurements of EMF 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 (v) In a potentiometer experiment, the balancing length is 8 m, when the two cells El and E2 are joined in series. When the two cells are connected in opposition the balancing length is 4 m. The ratio of the e. m. f. of two cells (El/E2) is
(a) 1: 2 (b) 2: 1 (c) 1: 3 (d) 3: 1
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CBSE 12th Standard Physics Subject Current Electricity Case Study Questions 2021 Answer Keys
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(I) (b): \(q=10^{6} \times 1.6 \times 10^{-19} \mathrm{C}=1.6 \times 10^{-13} \mathrm{C}\)
t = 10-3 s
\(I=\frac{q}{t}=\frac{1.6 \times 10^{-13}}{10^{-3}}=1.6 \times 10^{-10} \mathrm{~A}\)
(ii) (C): C S-1
(iii) (C): The current flowing through a conductor of non-uniform cross-section remain same in the whole of the conductor.
(iv) (a): When a constant current is flowing through a conductor of non-uniform cross-section, electron density does not depend upon the area of cross section, while current density, drift velocity and electric field all vary inversely with area of cross-section.
(v) (a): Given, I = 40 A ;A = 10m2
\(\therefore\) Current density, \(J=\frac{I}{A} \text { or } J=\frac{40}{10}=4 \mathrm{~A} / \mathrm{m}^{2}\) -
(i) (b)
(ii) (a): \(R=\frac{V}{I}=\frac{2}{10^{-6}}=2 \times 10^{6} \Omega\)
(iii) (d): Specific resistance depends upon the nature of material and is independent of mass and dimensions of the material
(iv) (a)
(v) (d): l = 1.0 m; D = 0.4 mm = 4 x 10-4m
\(R=2 \Omega\)
\(A=\frac{\pi D^{2}}{4}=\frac{\pi \times\left(4 \times 10^{-4}\right)^{2}}{4}=4 \pi \times 10^{-8} \mathrm{~m}^{2}\)
Now, \(\rho=\frac{R A}{l}=\frac{2 \times 4 \pi \times 10^{-8}}{1}=2.55 \times 10^{-7} \Omega \mathrm{m}\) -
(i) (b)
(ii) (b): As \(I=\frac{\varepsilon}{R+r}\)
In first case,\(I=0.5 \mathrm{~A} ; R=12 \Omega\)
\(0.5=\frac{\varepsilon}{12+r} \Rightarrow \varepsilon=6.0+0.5 r\) ....(i)
In second case \(I=0.25 \mathrm{~A} ; R=25 \Omega\)
\(\varepsilon=6.25+0.25 r\) ...(ii)
From equation (i) and (ii), r = \(1 \Omega\)
(iii) (b)
(iv) (a): Current in the circuit \(I=\frac{E}{R+r}\)
Power delivered to the resistance R is
\(P=I^{2} R=\frac{E^{2} R}{(R+r)^{2}}\)
It is maximum when \(\frac{d P}{d R}=0\)
\(\frac{d P}{d R}=E^{2}\left[\frac{(r+R)^{2}-2 R(r+R)}{(r+R)^{4}}\right]=0\)
\(\text { or } \quad(r+R)^{2}=2 R(r+R) \text { or } R=r\)
(v) (b): For first case, \(\frac{\varepsilon}{R+r}=\frac{10}{R}\) ...(i)
For second case, \(\frac{\varepsilon}{5 R+r}=\frac{30}{5 R}\)
Dividing (i) by (ii), we get r = 5R
From (i), \(\frac{E}{R+5 R}=\frac{10}{R}\)
E = 60 V -
(i) (d)
(ii) (c): For parallel combination of n celis, \(\varepsilon_{e q}=\varepsilon\)
(iii) (d): \(I=\frac{m E}{m R+r}\) m= number of cells = 4
\(E=2 \mathrm{~V}, R=2 \Omega, r=1 \Omega\)
\(I=\frac{8}{8+1}=\frac{8}{9}=0.888 \mathrm{~A}\)
(iv) (b)
(v) (c): Let two cells of emf's E1 and E2 and of internal resistance r1 and r2 respectively are connected in parallel
The equivalent emf is given by
\(\varepsilon_{\mathrm{eq}}=\frac{\varepsilon_{1} r_{2}+\varepsilon_{2} r_{1}}{r_{1}+r_{2}}\)...(I)
The equivalent internal resistance is given by
\(\frac{1}{r_{\mathrm{eq}}}=\frac{1}{r_{1}}+\frac{1}{r_{2}} \quad \text { or } \quad r_{\mathrm{eq}}=\frac{r_{1} r_{2}}{r_{1}+r_{2}}\)
Let us consider, two cells connected in parallel of same emf E and same internal resistance r.
From equatio. n (i), we get \(\varepsilon_{\mathrm{eq}}=\frac{\varepsilon r+\varepsilon r}{r+r}=\varepsilon\)
From equation (ii), we get
\(r_{\mathrm{eq}}=\frac{r^{2}}{r+r}=\frac{r}{2}\) -
(i) (b)
(ii) (a): Given, \(I=1.0 \mathrm{~mA}=10^{-3} \mathrm{~A} ; R=4 \Omega ; L=4 \mathrm{~m}\)
Potential drop across potentiometer wire
V = IR = 10-3 X 4 V
Potential gradient, \(k=\frac{V}{L}=\frac{4 \times 10^{-3}}{4}\)
= 10-3 V m-1
(iii) (a)
(iv) (b): A potentiometer is an accurate and versatile device to make electrical measurements of EMF because the method involves a condition of no current flow through the galvanometer. It can be used to
measure potential difference, internal resistance of a cell and compare EMF's of two sources.
(v) (d):\(\frac{E_{1}}{E_{2}}=\frac{l_{1}+l_{2}}{l_{1}-l_{2}}=\frac{8+4}{8-4}=\frac{12}{4}=\frac{3}{1}\)
