Solutions-Class-12-Physics-Chapter-9-Current Electricity-MSBSHSE

(A) charge

(D) neither cell nor auxiliary battery

(C) 2 Ω

(D) 30 Ω

(D) 3A

(A) 2 Ω

The potentiometer is a device used for accurate measurement of potential difference as well as the emf of a cell.

• It does not draw any current from the circuit at the null point. Thus, it acts as an ideal voltmeter and it can be used to determine the internal resistance of a cell.
• It consist of a long uniform wire AB of length L, stretched on a wooden board. A cell of stable emf (E), with a plug key K in series, is connected across AB as shown in the following figure.

Potential gradient is defined as the potential difference (the fall of potential from the high potential end) per unit length of the wire.

Sliding the jockey on the potentiometer wire decreases the cross sectional area of the wire and thereby affects the fall of potential along the wire.

This affects the potentiometer readings. Hence, the jockey should not be slided along the potentiometer wire.

Kirchhoff’s laws are applicable to both AC and DC circuits (networks). For AC circuits with different loads, (e.g. a combination of a resistor and a capacitor, the instantaneous values for current and voltage are considered for addition.

The value of unknown resistance X, may not be accurate due to non-uniformity of the bridge wire and development of contact resistance at the ends of the wire.

To minimize these errors, the value of R is so adjusted that the null point is obtained in the middle one-third of the wire (between 34 cm and 66 cm) so that the percentage errors in the measurement of l_X and l_R are minimum and nearly the same.

Errors are almost unavoidable but can be minimized considerably as follows :

(1) The value of unknown resistance X, may not be accurate due to non-uniformity of the bridge wire and development of contact resistance at the ends of the wire.

To minimize the errors

• The value of R is so adjusted that the null point is obtained to middle one third of the wire (between 34 cm and 66 cm) so that percentage error in the measurement of l_X and l_R are minimum and nearly the same.

(2) The ends of the wire are soldered to the metallic strip where contact resistance is developed, which is not taken into account.

To minimize the errors

• The experiment is repeated by interchanging the positions of unknown resistance X and known resistance box R.

Potential gradient : Potential gradient is defined as the potential difference (the fall of potential from the high potential end) per unit length of the wire.

Consider A potentiometer consists of a long wire AB of length L and resistance R having uniform cross sectional area A. (see above Fig.) A cell of emf E having internal resistance r is connected across AB as shown in the Fig. When the circuit is switched on, current I passes through the wire.

Current through AB, I = \(\frac{E}{R+r}\)

Potential difference across AB. V_AB = IR

∴ V_AB = \(\frac{ER}{R+r}\)

The potential difference (the fall of potential from the high potential end) per unit length of the wire,

\(\frac{V_{AB}}{L}=\frac{ER}{R+r}\)

As long as E and r remain constant, \(\frac{V_{AB}}{L}\) will remain constant.

\(\frac{V_{AB}}{L}\) is known as potential gradient along AB and is denoted by K.

Thus the potential gradient is calculated by measuring the potential difference between ends of the potentiometer wire and dividing it by the length of the wire.

Let C be any point on the wire between A and B and AC = l = length of the wire between A and C.

Then V_AC = Kl

 V_AC ∝ l as K is constant in a particular case.

Thus, the potential difference across any length of the potentiometer wire is directly proportional to that length. This is the principle of the potentiometer.

The potential gradient depends upon the potential difference between the ends of the wire and the length of the wire.

• A voltmeter should ideally have an infinite resistance so that it does not draw any current from the circuit. However a voltmeter cannot be designed to have infinite resistance.
• A potentiometer does not draw any current from the circuit at the null point. Therefore, it gives more accurate measurement. Thus, it acts as an ideal voltmeter.

. Some practical applications of potentiometer are given below.

• Voltage divider : The potentiometer can be used as a voltage divider to change the output voltage of a voltage supply.
• Audio control : Sliding potentiometers are commonly used in modern low-power audio systems as audio control devices. Both sliding (faders) and rotary potentiometers (knobs) are regularly used for frequency attenuation, loudness control and for controlling different characteristics of audio signals.
• Potentiometer as a sensor : If the slider of a potentiometer is connected to the moving part of a machine, it can work as a motion sensor. A small displacement of the moving part causes changes in potential which is further amplified using an amplifier circuit. The potential difference is calibrated in terms of the displacement of the moving part.
• To measure the emf (for this, the emf of the standard cell and potential gradient must be known).
• To compare the emf’s of two cells.
• To determine the internal resistance of a cell.

Disadvantages of a potentiometer :

• The use of a potentiometer is an indirect measurement method while a voltmeter is a direct reading instrument.
• A potentiometer is unwieldy while a voltmeter is portable.
• Unlike a voltmeter, the use of a potentiometer in measuring an unknown emf requires a standard source of emf and calibration.

(i) On increasing the current through the potentiometer wire, the potential gradient along the wire will increase. Hence, the position of zero deflection will occur at a shorter length.

(ii) On decreasing the current through the potentiometer wire, the potential gradient along the wire will decrease. Hence, the position of zero deflection will occur at a longer length.

Wheatstone’s network and condition of balance :

Wheatstone’s network or bridge is a circuit for indirect measurement of resistance by null comparison method by comparing it with a standard known resistance.

It consists of four resistors with resistances P, Q, R and S arranged in the form of a quadrilateral ABCD.

A cell (E) with a plug key (K) in series is connected across one diagonal AC and a galvanometer (G) across the other diagonal BD as shown in the following figure.

With the key K closed, currents pass through the resistors and the galvanometer. One or more of the resistances is adjusted until no deflection in the galvanometer can be detected. The bridge is then said to be balanced.

Let I be the current drawn from the cell. At junction A, it divides into a current I₁ through P and a current I₂ through S.

I = I₁ + I₂ (by Kirchhoff’s first law).

At junction B, current I_g flows through the galvanometer and current I₁ − I_g flows through Q.

At junction D, I₂ and I_g combine. Hence, current I₂ + I_g flows through R from D to C. At junction C, I₁ − I_g and I₂ + I_g combine.

Hence, current I₁ + I₂(=I) leaves junction C.

Applying Kirchhoff’s voltage law to loop ABDA in a clockwise sense, we get,

−I₁P − I_gG + I₂S = 0  …….(1)

where G is the resistance of the galvanometer.

Applying Kirchhoff’s voltage law to loop BCDB in a clockwise sense, we get,

− (I₁ − I_g)Q + (I₂ + I_g)R + I_gG = 0  …….(2)

When I_g = 0, the bridge (network) is said to be balanced.

In that case, from Eqs. (1) and (2), we get,

I₁P = I₂S  ……(3)

and I₁Q = I₂R   ……(4)

From Eqs. (3) and (4), we get,

P/Q = S/R

This is the condition of balance.

A metre bridge consists of a rectangular wooden board with two L-shaped thick metallic strips fixed along its three edges. A single thick metallic strip separates two L-shaped strips. A wire of length one metre and uniform cross section is stretched on a metre scale fixed on the wooden board. The ends of the wire are fixed to the L-shaped metallic strips.

Determining unknown resistance using meter bridge :

An unknown resistance X is connected in the left gap and a resistance box R is connected in the right gap as shown in Fig. One end of a centre-zero galvanometer (G) is connected to terminal C and the other end is connected to a pencil jockey (J).

A cell (E) of emf E, plug key (K) and rheostat (Rh) are connected in series between points A and B.

Working : Keeping a suitable resistance (R) in the resistance box, key K is closed to pass a current through the circuit. The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows zero deflection. The bridge is then balanced and point D is called the null point and the method is called as null deflection method.

The distances I_X and I_R of the null point from the two ends of the wire are measured.

According to the principle of Wheatstone’s network,

\(\frac{X}{R}=\frac{\text{resistance of the wire of length}I_X(R_{AD})}{\text{resistance of the wire of length}I_R(R_{DB})}\)

∴ \(\frac{X}{R}=\frac{R_{AD}}{R_{DB}}\)

Now, R =  where l is the length of the wire, ρ is the resistivity of the material of the wire and A is the area of cross section of the wire.

∴ R_AD = \(ρ\frac{l_X}{A}\) and R_DB = \(ρ\frac{l_R}{A}\)

∴ \(\frac{X}{R}=\frac{R_{AD}}{R_{DB}}\) =\(\frac{ρl_X/A}{ρl_R/A}=\frac{l_X}{l_R}\)

∴ X = \(\frac{l_X}{l_R}\)xR

As R, l_X and l_R are known, the unknown resistance X can be calculated.

Kelvin’s method :

Circuit :The metre bridge circuit for Kelvin’s method of determination of the resistance of a galvanometer is shown in below Fig. The gaivanometer whose resistance G is to be determined, is connected in one gap of the metre bridge-

A resistance box providing a variable known resistance R is connected in the other gap. The junction B of the galvanometer and the resistance box is connected directly to a pencil jockey. A cell of emf E, a key (K) and a rheostat (Rh) are connected across AC.

Working :

Keeping a suitable resistance R in the resistance box and maximum resistance in the rheostat, key K is closed to pass the current.

The rheostat resistance is slowly reduced such that the galvanometer shows about 2/3rd of the full—scale deflection.

On tapping the jockey at end-points A and C, the galvanometer deflection should change to opposite sides of the initial deflection. Only then will there be a point D on the wire which is equipotential with point B.

The jockey is tapped along the wire to locate the equipotential point D when the galvanometer shows no change in deflection.

Point D is now called the balance point and Kelvin's method is thus an equal deflection method. At this balanced condition,

\(\frac{G}{R}\)=\(\frac{\text{resistance of the wire of length}I_G}{\text{resistance of the wire of length}I_R}\)

where I_G ≡ the length of the wire opposite to the galvanometer,

I_R ≡ the length of the wire opposite to the resistance box.

If λ ≡ the resistance per unit length of the wire,

\(\frac{G}{R}\)=\(\frac{λI_G}{λI_R}\)=\(\frac{I_G}{I_R}\)

∴ G = R\(\frac{I_G}{I_R}\)

The quantities on the right hand side are known, so that G can be calculated.

A battery of stable emf E is used to set up a potential gradient V/L along a potentiometer wire, where V ≡ potential difference across the length L of the wire. The positive terminals of the cells, whose emf’s (E₁ and E₂) are to be compared, are connected to the high potential terminal A. The negative terminals of the cells are connected to a galvanometer G through a two-way key. The other terminal of the galvanometer is connected to a pencil jockey. The emf E should be greater than both the emf’s E₁ and E₂.

Connecting point P to C, the cell with emf E₁ is brought into the circuit. The jockey is tapped along the wire to locate the null point D at a distance l₁ from A. Then,

E₁ = l₁(V/ L)

Now, without changing the potential gradient (i.e., without changing the rheostat setting) point Q (instead of P) is connected to C, bringing the cell with emf E₂ into the circuit. Let its null point D’ be at a distance l₂ from A, so that

E₂ = l₂(V/ L)

∴ E₁/E₂ = l₁/l₂

Hence, by measuring the corresponding null lengths l₁ and l₂, E₁/E₂ can be calculated. The experiment is repeated for different potential gradients using the rheostat.

A battery of stable emf E is used to set up a potential gradient V/L, along the potentiometer wire, where V ≡ potential difference across length L of the wire.

The positive terminal of the cell 1 is connected to the higher potential terminal A of the potentiometer; the negative terminal is connected to the galvanometer G through the reversing key. The other terminal of the galvanometer is connected to a pencil jockey.

The cell 2 is connected across the remaining two opposite terminals of the reversing key. The other terminal of the galvanometer is connected to a pencil jockey.

The emf E₁ should be greater than the emf E₂; this can be adjusted by trial and error.

Two plugs are inserted in the reversing key in positions 1—1. Here, the two cells assist each other so that the net emf is E₁ + E₂. The jockey is tapped along the wire to locate the null point D. If the null point is a distance l₁ from A,

E₁ + E₂ = l₁ (V/L)

For the same potential gradient (without changing the rheostat setting), the plugs are now inserted into position 2—2. (instead of 1—1).

The emf E₂ then opposes E₁ and the net emf is E₁ − E₂.

The new null point D’ is, say, a distance l₂ from A and

E₁ − E₂ = l₂ (V/L)

∴ \(\frac{E_1+E_2}{E_1-E_2}=\frac{l_1}{l_2}\)

∴ \(\frac{E_1}{E_2}=\frac{l_1+l_2}{l_1-l_2}\)

Here, the emf E should be greater than E₁ + E₂

The experiment is repeated for different potential gradients using the rheostat.

Principle : A cell of emf E and internal resistance r, which is connected to an external resistance R, has its terminal potential difference V less than its emf E.

If I is the corresponding current,

\(\frac{E}{V}=\frac{I(R+r)}{IR}\)=\(\frac{R+r}{R}\)

= \(1+\frac{r}{R}\)  ( when R → ∞, v→ E)

∴ r = \(\frac{E-V}{V}R\)

Working : A battery of stable emf E’ is used to set up a potential gradient V/L, along the potentiometer wire, where V ≡ potential difference across the length L of the wire.

The negative terminal is connected through a centre-zero galvanometer G to a pencil jockey. A resistance box R with a plug key K in series is connected across the cell.

Firstly, key K is kept open; then, effectively, R = ∞. The jockey is tapped on the potentiometer wire to locate the null point D. Let the null length

AD = l₁, so that

E = (V_AB/L)l₁

With the same potential gradient, and a small resistance R in the resistance box, ‘key K is closed.

The new null length AD’ = l₂ for the terminal potential difference V is found :

V = (V_AB /L)l₂

∴ E/V = \(\frac{l_1}{l_2}\)   ∴\(\frac{E-V}{V}\)=\(\frac{l_1-l_2}{l_2}\)=\(\frac{l_1}{l_2}-1\)

Now  r = \(\frac{E-V}{V}R\)

∴ r = \(R(\frac{l_1}{l_2}-1)\)

R, l₁ and l₂ being known, r can be calculated. The experiment is repeated with different potential gradients using the rheostat or with different values of R.

The internal resistance of a cell depends on :

• Nature of the electrolyte : The greater the conductivity of the electrolyte, the lower is the internal resistance of the cell.
• Separation between the electrodes : The larger the separation between the electrodes of the cell, the higher is the internal resistance of the cell. This is because the ions have to cover a greater distance before reaching an electrode.
• Nature of the electrodes.
• The internal resistance is inversely proportional to the common area of the electrodes dipping in the electrolyte.

Let I₁ and I₂ be the currents through the two branches as shown in Fig.

The current through the 2 Ω resistance will be (I₁ + I₂) [Kirchhoff's current law].

Applying Kirchhoff’s voltage law to loop ABCDEFA, we get,

−2(I₁ + I₂) − 1(I₁) + 4 = 0

 3I₁ + 2I₂ = 4  ….(1)

Applying Kirchh0ff’s voltage law to loop BCDEB we get,

−2(I₁ + I₂) − 1(I₂) + 1 = 0

2I₁ + 3I₂ = 1  …..(2)

Multiplying Eq. (1) by 2 and Eq. (2) by 3, we get,

6I₁ + 4I₂ = 8  ……(3)

and 6I₁ + 9I₂ = 3  ……(4)

Subtracting Eq. (4) from Eq. (3), we get,

−5I₂ = 5

∴ I₂ = −1A

The minus sign shows that the direction of current I₂ is opposite to that assumed. Substituting this value of I₂ in Eq. (1), we get,

3I₁ + 2(−1) = 4

∴ 3 I₁ = 4 + 2 = 6

∴ I₁ = 2A

Current through the 2 Ω resistance = I₁ + I₂ = 2 – 1 = 1A It is in the direction as shown in figure.

Let I₁ and I₂ be the currents through the two branches as shown in Fig.

The current through the 5 Ω resistor will be I₁ + I₂ [Kirchhoff’s current law].

Applying Kirchhoff’s voltage law to loop ABCDEFA, we get,

−5(I₁ + I₂) − I₁ + 1.5 = 0

 6I₁ + 5I₂ = 1.5     ….(1)

Applying Kirchh0ff’s voltage law to loop BCDEB we get,

−5(I₁ + I₂) − 2I₂ + 2 = 0

5I₁ +7I₂ = 2     ….(2)

Multiplying Eq. (1) by 5 and Eq. (2) by 6, we

30I₁ + 25I₂ = 7.5    

and 30I₁ + 42I₂ = 12

Subtracting Eq. (3) from Eq. (4), we get,

17I₂ = 4.5

∴ I₂ = 4.5/17 A

Substituting this value of I₂ in Eq. (1), we get,

6I₁ + 5(4.5/17) = 1.5      (3)

∴ 6I₁ + (22.5/17) = 1.5

∴ 6I₁ = \(1.5-\frac{22.5}{17} = \frac{28.5-22.5}{17}\) 

= \(\frac{3}{17}\)

∴ I₁ = \(\frac{3}{17×6}\) = \(\frac{0.5}{17}\)A

Current through the 5 Ω resistance (external resistance)

=  I₁ + I₂ = \(\frac{0.5}{17}+\frac{4.5}{17}\) = \(\frac{5}{17}\)A

Given : E =2V, r = 10 Ω , R =30 Ω

The voltmeter reading,

V = IR = \((\frac{E}{R+r})R\)

            =  \((\frac{2}{30+10})30\)                        

            = \(\frac{3}{2}\) = 1.5 V

Figure shows the electrical network. For resistances 10 Ω, 12 Ω and 15 Ω connected in parallel the equivalent resistance (R_p) is given by,

\(\frac{1}{R_p}\)=\(\frac{1}{10}+\frac{1}{12}+\frac{1}{15}\) = \(\frac{6+5+4}{60}=\frac{1}{4}\)

∴ R_p = 4

For resistance R_P, 10 Ω, 12 Ω and 15 Ω connected in series, the equivalent resistance,

R_s = 4 + 10 + 12 + 15 = 41 Ω

Thus, the total resistance = R_s = 41 Ω

Now, V = IR_s

 4.1 = I x 41

∴ I = 0.1 A

The total resistance and current through the circuit are 41 Ω and 0.1 A respectively.

Given : L = 1.5 m, R = 10 Ω, E = 4V, r = 5 Ω.

K = \(\frac{ER}{(R+r)L}\)

∴ K = \(\frac{4×10}{(10+5)1.5}=\frac{40}{22.5}\frac{V}{m}=\frac{40}{2250}\frac{V}{cm}\) = 0.0178 V/cm

The potential drop per centimeter of the wire is 0.0178 V/cm

Given : l₁ = 2.7 m (cells assisting), l₂ = 0.3 m (cells opposing)

E₁ + E₂ = Kl₁ and E₁ − E₂ = Kl₂

∴ \(\frac{E_1+E_2}{E_1-E_2}=\frac{Kl_1}{Kl_2}\)

∴ \(\frac{E_1}{E_2}=\frac{l_1+l_2}{l_1-l_2}=\frac{2.7+0.3}{2.7-0.3}=\frac{3}{2.4}\)= 1.25

The ratio of the emf's of the two cells is 1.25

Given : R = 10 Ω, l₁ =120 cm, l₂ = 120—20 = 100 cm

r = R\((\frac{l_1-l_2}{l_2})\) = \(10(\frac{120-100}{100})\)

   = 2 Ω

The internal resistance of the cell is 2 Ω.

Given : K = 5 x 10⁻³ V/m, L = 216 cm = 216 x 10⁻² m

E = KL

∴ E = 5 x 10⁻³  x 216 x 10⁻²

= 1080 x 10⁻⁵

= 0.01080 V

The emf of the cell is 0.01080 volt.

Given : R = 8 Ω, L = 8m, E = 2V, K = 1 μV/mm = 1 x 10⁻³ V/m

K = V/L = \(\frac{ER}{(R+R_B)L}\), where R_B is the resistance in the box.

∴ 1 x 10⁻³ = \(\frac{2×8}{(8+R_B)8}\)

∴ 8 + R_B = 2/10⁻³ = 2 x 10³

∴  R_B = 2000 − 8

        = 1992 Ω

Applying Kirchhoff’s voltage law to loop FGHF, we get,

−10I₁ − 10(I₁ − I₂) + 10(I − I₁) + 10(I − I₁) =0

 −10I₁ − 10I₁ + 10I₂ + 10I − 10I₁ + 10I – 10I₁ = 0

 20I − 40I₁ + 10I₂ = 0

 2I – 4I₁ + I₂ = 0    …….(1)

Applying Kirchhoffs voltage law to loop GABHG, we get,

−10I₂ − 10I₂ + 10(I − I₂)+ 10(I₁ − I₂) = 0

 – 20I₂ + 10I – 10 I₂ + 10I₁ – 10I₂ = 0

 10I + 10I₁ – 40I₂ = 0

 I + I₁ – 4I₂ = 0         ……(2)

Applying Kirchhoff’s voltage law to loop EFHBCDE, we get,

–10(I − I₁) − 10(I − I₁) − 10(I − I₂) + E =0

 − 10I + 10 I₁ − 10I+ 10 I₁ − 10I+ 10 I₂ + E =0

 E = 30I – 20I₁ − 10I₂      …..(3)

From Eq. (1), we get, I₂ = 4I₁ – 2I       …..(4)

From Eqs. (2) and (4), we get,

I + I₁ − 4(4I₁ − 2I) = 0

 I + I₁ – 16I₁ + 8I = 0

I + I₁ − 4(4I₁ – 2I) = 0

 I + I₁ – 16I₁ + 8I = 0

9I = 15I₁  ∴ I₁ = \(\frac{9}{15}I = \frac{3}{5}I\)    …..(5)

From Eqs. (4) and (5), we get,

I₂ = 4\((\frac{3}{5}I)\) − 2I = \(\frac{12}{5}I\) − 2I

= \(\frac{12I-10I}{5}\) = \(\frac{2}{5}I\)   …..(6)

From Eqs. (3), (5) and (6), we get

E = 30I − 20\((\frac{3}{5}I)\)  − 10\((\frac{2}{5}I)\)  = 30I − 12I – 4I 

∴ E = 14 I

If R is the equivalent resistance between E and C,

E = RI

∴ R = 14 Ω

Given : E = 2V, r = 20 Ω , R = 100 Ω

The voltmeter reading,

V = IR = \((\frac{E}{R+r})R\)

= \((\frac{2}{100+20})100=\frac{200}{120}\) = 1.667 V