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CBSE 12th Standard Physics Subject Current Electricity Chapter Case Study Questions 2021

Published on: 21/05/2021

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Std: 12th Standard CBSE - Physics

chapter: - Case study Questions

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Questions:

Case Study Questions

3 x 4 = 12

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.

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

\(\text { (a) } 1.57 \times 10^{-6} \Omega \mathrm{m}\)

\(\text { (b) } 5.25 \times 10^{-7} \Omega \mathrm{m}\)

\(\text { (c) } 7.12 \times 10^{-5} \Omega \mathrm{m}\)

\(\text { (d) } 2.55 \times 10^{-7} \Omega \mathrm{m}\)

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

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Answers Key:

Case Study Questions

3 x 4 = 12

(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 r₁and 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): \(R=\frac{V}{I}=\frac{2}{10^{-6}}=2 \times 10^{6} \Omega\)
(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): \(q=10^{6} \times 1.6 \times 10^{-19} \mathrm{C}=1.6 \times 10^{-13} \mathrm{C}\)
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, \(J=\frac{I}{A} \text { or } J=\frac{40}{10}=4 \mathrm{~A} / \mathrm{m}^{2}\)