Physics 5 Marks Questions

1. Equivalent resistance, $\text{R}=\frac{11}{3}\Omega$

Consider the followinq combination of the resistors.

Equivalent resistance of the circuit is given by,

$\text{R}=\frac{2\times1}{2+1}+3=\frac{2}{3}+3=\frac{11}{3}\Omega$

2. Equivalent resistance, $\text{R}=\frac{11}{5}\Omega$

Consider the following combtnatlon of the resistors.

Equivalent resistance of the circuit is given by,

$\text{R}=\frac{2\times3}{2+3}+1=\frac{6}{5}+1=\frac{11}{5}\Omega$

3. Equivalent resistance, $\text{R}=6\ \Omega$

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by the sum.

$\text{R}=1+2+3=6\ \Omega$

4. Equivalent resistance, $\text{R}=\frac{6}{11}\Omega$

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by,

$\text{R}=\frac{1\times2\times3}{1\times2+2\times3+3\times1}=\frac{6}{11}\Omega$

1. Constant emf of the given standard cell, E₁ = 1.02 V

Balance point on the wire, l₁ = 67.3 cm

A cell of unknown emf, $\varepsilon,$ replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm

The relation connecting emf and balance point is,

$\frac{\text{E}_1}{l_1}=\frac{\varepsilon}{l}$

$\varepsilon=\frac{l}{l_1}\times\text{E}_1$

$=\frac{82.3}{67.3}\times1.02=1.247\ \text{V}$.

The value of unknown emf is 1.247 V.

2. The purpose of using the high resistance of $600\ \text{k}\Omega$ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
3. The balance point is not affected by the presence of high resistance.
4. The point is not affected by the internal resistance of the driver cell.
5. The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.
6. The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

1. It can be observed from the given circuit that in the first small loop, two resistors of resistance $1\ \Omega$ each are connected in series.

Hence, their equivalent resistance $=(1+1)=2\ \Omega$

It can also be observed that two resistors of resistance $2\ \Omega$ each are connected in series.

Hence, their equivalent resistance $=(2+2)=4\ \Omega$

Therefore, the circuit can be redrawn as

It can be observed that $2\ \Omega\ \text{and}\ 4\ \Omega$ resistors are connected in parallel in all the four loops. Hence, equivalent resistance (R') of each loop is given by,

$\text{R}=\frac{2\times4}{2+4}=\frac{8}{6}=\frac{4}{3}\Omega$

The circuit reduces to,

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is $\frac{4}{3}\times4=\frac{16}{3}\Omega$

2. It can be observed from the given circuit that five resistors of resistance R each are connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R + R

= 5 R

$V=\varepsilon-Ir$

When current is zero (I = 0), $\text{V}=\in$

And when V = 0, $\text{I}=\text{I}_0,\text{ }r=\frac{\in}{I_0}$

$\text{V}=\text{V}(\text{B}_1)-\text{V}(\text{B}_2)=\varepsilon_1-I_1r_1$

$\text{V}=\text{V}(\text{B}_1)-\text{V}(\text{B}_2)=\varepsilon_2-I_2r_2$

$I=I_1+I_2$

$=​​​​​\frac{​\varepsilon_1-\text{V}}{r_1}+\frac{​\varepsilon_2-\text{V}}{r_2}=\Big(\frac{​\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}\Big)-\text{V}\Big(\frac{1}{r_1}+\frac{1}{r_2}\Big)$

$\text{V}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}-I\frac{r_1r_2}{r_1+r_2}$

On comparing with

$\text{V}=\varepsilon_{eq}-Ir_{eq}$

we get

$\varepsilon_{eq}=\frac{\varepsilon_1r_2+\varepsilon_2r_1}{r_1+r_2}$

$r_{eq}=\frac{r_1r_2}{r_1+r_2}$

Alternate Answer

A student may write the last two results in the following form.

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}$

$\frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2}$

1. The potential difference across any length of wire is directly proportional to the length provided current and area of cross section are constant i.e., 𝐸 𝑙 = 𝜙𝑙 where 𝜙 is the potential drop per unit length.

It can be made more sensitive by decreasing current in the main circuit/decreasing potential gradient/increasing resistance put in series with the potentiometer wire.

2. Potentiometer B

Has smaller value of $\text{ }^V/_l$ (slope/potential gradient).

2. In series, the current remains the same.

$I=neA_1V_{d1}=neA_2V_{d2}$

$\therefore\frac{V_{d1}}{V_{d2}}=\frac{A_2}{A_1}$

In parallel potential difference is same but currents are different.

$V=I_1R_1=neA_1V_{d1}\frac{\rho l}{A_1}=ne\rho V_{d1}l$

Similarly, $V=I_2R_2=ne\rho V_{d2}l$

$I_1R_1=I_2R_2$

$\therefore\frac{V_{d1}}{V_{d2}}=1$

1. Current through AB

$I=\frac{\varepsilon_1}{R+R_{AB}}=\frac{2}{R+15}$

P.D. across AB, $V_{AB}=IR_{AB}=\big(\frac{2}{R+15}\big).15$

Potential gradient $k=\frac{V_{AB}}{AB}=\frac{30}{(R+15)\times100}=\frac{0.3}{R+15}$

Balance length for cell E₂ (= 75mV), $l=\frac{E_2}{k}$

$\Rightarrow30=\frac{75\times10^{-3}(R+15)}{0.3}$

$\Rightarrow\frac{9\times10^3}{75}=R+15$

$\Rightarrow R=105\Omega$

2. A potentiometer is preferred over a voltmeter because potentiometer does not draw current for any measurement unlike a voltmeter.

(Alternatively, Potentiometer compares the emf values while the voltmeter would only compare the terminal p.d.’s of the two cells.)

1. Let an electric field E be applied the conductor. Acceleration of each electron is

$a=-\frac{eE}{m}$

Velocity gained by the electron

$v=-\frac{eE}{m}t$

Let the conductor contain n electrons per unit volume. The average value of time $'t'$, between their successive collisions, is the relaxation time,$'\tau'$.

Hence average drift velocity $v_d=\frac{-eE}{m}\tau$

The amount of charge, crossing area A, in time $\triangle t$, is

$\equiv neAv_d\triangle t=I\triangle t$

Substituting the value of $v_d$, we get

$I\triangle t=neA\bigg(\frac{eE\tau}{m}\bigg)\triangle t$

$\therefore\text{ }I=\bigg(\frac{e^2A\tau n}{m}\bigg)E=\sigma E,\bigg(\sigma=\frac{e^2\tau n}{m} \text{is the conductivity}\bigg)$

But I = JA, where J is the current density

$\Rightarrow\text{J}=\big(\frac{e^2\tau n}{m}\big)E$

$\Rightarrow\text{J}=\sigma E$

This is Ohm’s law

2. Electric current well remain constant in the wire. All other quantities, depend on the cross sectional area of the wire.

1. Average velocity acquired by the electrons in the conductor in the presence of external electric field.

Alternate Answer

$[\text{V}_{d} =\frac{-eE\tau}{\text{m}}$ where $\tau$is the relaxation time.]

2. $\text{V}_{d} =\frac{-eE\tau}{\text{m}}$ 

We have $\text{E} = - \frac{\text{V}}{\ell'}$

where V is potential difference across the length $\ell$

of the conductor

$\text{V}_{d} = \frac{\text{eV}\tau}{\text{m}\ell}$ Current flowing I = neAv_d

I = neA$\text{V}_{d}\frac{\text{eV}\tau}{\text{ml}} = \frac{ne^{2}\text{AV}\tau}{\text{ml}}$

$\frac{\text{I}}{\text{V}} = \frac{\text{ne}^{2}A\tau}{ml} = \frac{1}{\text{R}}$ . . . . . (i)

Also,  $\text{R} = \rho\frac{\ell}{\text{A}}$ . . . . (ii) 

Comparing (i) and (ii)

$\rho = \frac{m}{ne^{2}\tau}$

Resistivity of the material of a conductor depends on the relaxation time, i.e., temperature and the number density of electrons.

3. Because constantan and manganin show very weak dependence of resistivity on temperature.

1. Junction Rule: At any Junction, the sum of currents, entering the junction, is equal to the sum of currents leaving the junction.

Loop Rule: The Algebraic sum, of changes in potential, around any closed loop involving resistors and cells, in the loop is zero.

$\sum(\triangle V)=0$

Justification: The first law is in accord with the law of conservation of charge.

The Second law is in accord with the law of conservation of energy.

2. Equivalent resistance of the loop

$\text{R}=\frac{r}{3}$

Hence current drawn from the cell

$\text{I}=\frac{E}{\frac{r}{3}+r}=\frac{3E}{4r}$

Power consumed $\text{P}=I^2(\frac{r}{3})$

$=\frac{9E^2}{16r^2}\times\frac{4r}{3}=\frac{3E^2}{4r}$

1. Working Principle of Potentiometer:

Principle. Consider a long resistance wire AB of uniform cross-section. Its one end A is connected to the positive terminal of battery B₁ whose negative terminal is connected to the other end B of the wire through key K and a rheostat (Rh). The battery B₁ connected in circuit is called the driver battery and this circuit is called the primary circuit. By the help of this circuit a definite potential difference is applied across the wire AB; the potential falls continuously along the wire from A to B. The fall of potential per unit length of wire is called the potential gradient. It is denoted by ‘k’. A cell e is connected such that its positive terminal is connected to end A and the negative terminal to a jockey J through the galvanometer G. This circuit is called the secondary circuit.

In primary circuit the rheostat (Rh) is so adjusted that the deflection in galvanometer is on one side when jockey is touched on wire at point A and on the other side when jockey is touched on wire at point B.

The jockey is moved and touched to the potentiometer wire and the position is found where galvanometer gives no deflection. Such a point P is called null deflection point.

V_AB is the potential difference between points A and B and L metre be the length of wire, then the potential gradient

$\text{k} =\frac{\text{V}_{AB}}{\text{L}}$

If the length of wire AP in the null deflection position be l, then the potential difference between points A and P,

V_AP = kl

$\therefore\text{The emf of cell,}\varepsilon =\text{V}_{AP} = kl$

In this way the emf of a cell may be determined by a potentiometer.

Comparison of emf’s of two cells: First of all the ends of potentiometer are connected to a battery B₁, key K and rheostat Rh such that the positive terminal of battery B₁ is connected to end A of the wire. This completes the primary circuit. Now the positive terminals of the cells C₁ and C₂ whose emfs are to be compared are connected to A and the negative terminals to the jockey J through a two-way key and a galvanometer (fig). This is the secondary circuit.

Method:

1. By closing key K, a potential difference is established and rheostat is so adjusted that when jockey J is made to touch at ends A and B of wire, the deflection in galvanometer is on both sides. Suppose in this position the potential gradient is k.

2. Now plug is inserted between the terminals 1 and 3 so that cell C₁ is included in the secondary circuit and jockey J is slided on the wire at P₁ (say) to obtain the null point. The distance of P₁ from A is measured. Suppose this length is l₁ i. e. AP₁ =l₁

_{$\therefore\text{The emf of cell C}_{1},\varepsilon = kl_{1}$ -  - - - -(i)}

3. Now plug is taken off between the terminals 1 and 3 and inserted in between the terminals 2 and 3 to bring cell C₂ in the circuit. Jockey is slided on wire and null deflection position P₂ is noted. Suppose distance of P₂ from A is l₂ i. e. AP₂ =l₂

$\therefore\text{The emf of cell C}_{2},\varepsilon_{2} = kl_{2}$ - - - - - -- (ii)

Dividing (i) by (ii), we get

$\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{l_{1}}{l_{2}}$ - - - - -  - - - -(iii)

Thus emf’s of cells may be compared. Out of these cells if one is standard cell, then the emf of other cell may be calculated.

2. Possible causes for one side deflection:
   1. If emf $\varepsilon_{1}$(or $\varepsilon_{2}$ ) is more than the emf driver cell (auxiliary battery), then we have one sided deflection.
   2. ​​​when the positive end of the potentiometer wire is connected to negative terminal of the cell whose emf is to be determined.

1. Kirchhoff’s Rule
   1. At any junction, the sum of the currents entering the junction is equal to the sum of the currents leaving the junction.
   2. The algebraic sum of the charges in potential around any closed loop involving resistors and cells in the loop is zero.

Condition of balance of a Wheatstone bridge: The circuit diagram of Wheatstone bridge is shown in fig.

P, Q, R and S are four resistance forming a closed bridge, called Wheatstone bridge. A battery is connected across A and C, while a galvanometer is connected between B and D. At  balance, there is no current in galvanometer.

Derivation of Formula: Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I₁ and I₂ . As there is no current in galvanometer in balanced state, current in resistances P and Q is I₁ and in resistances R and S it is I₂ .

Applying Kirchhoff’s I law at point A

I - I₁ - I₂ = 0 or I = I₁ + I₂ - - - - - - - - (i)

Applying Kirchhoff’s II law to closed mesh ABDA

- I₁P + I₂R = 0 or I₁P = I₂ R -  - - - - - - (ii)

Applying Kirchhoff’s II law to mesh BCDB

- I₁Q + I₂S = 0 or I₁Q = I₂S - - - - -- - (iii)

Dividing equation (ii) by (iii), we get

$\frac{\text{I}_{1}\text{P}}{\text{I}_{1}\text{Q}} = \frac{\text{I}_{2}\text{R}}{\text{I}_{2}\text{S}}\text{ or }\frac{\text{P}}{\text{Q}} = \frac{\text{R}}{\text{S}}$ - - - - - - -(iv)

This is the condition of balance of Wheatstone bridge.

For null point at D, balance length $\ell_{1}$ = 40 cm

So,  $\frac{\text{R}_{1}}{\text{R}_{2}} = \frac{\text{AD}}{\text{DC}} =\frac{40}{(100- 40)} = \frac{2}{3}$ - - - -- - (i)

If resistance 10 $\Omega$ is connected in series of R₁, then balance length AD' > AD i.e. balance point shifts by length ‘y’ towards C i.e., AD = 60 cm.

$\frac{\text{R}_{1} + 10}{\text{R}_{2}} = \frac{\text{AD}"}{\text{D}'\text{C}} = \frac{60}{100-60} = \frac{3}{2}$

$\frac{\text{R}_{1}}{\text{R}_{2}} + \frac{10}{\text{R}_{2}} =\frac{3}{2}$-  - - - - -- -- - -(ii)

From equations (1) and (2), we have

$\frac{2}{3} +\frac{10}{\text{R}_{2}} =\frac{3}{2}$

$\frac{10}{\text{R}_{2}} = \frac{3}{2} - \frac{2}{3} =\frac{9-4}{6} = \frac{5}{6}$

$\Rightarrow\text{R}_{2} = \frac{ 10\times6}{5} = 12\text{ ohm}$

From equation (1), we have

$\frac{\text{R}_{1}}{12} = \frac{2}{3}\Rightarrow\text{R}_{1} = \frac{12\times2}{3} =8\text{ ohm}.$

1. The equivalent circuit is shown in fig. It is a balanced Wheatstone bridge.

So, the resistance connected between C and D is ineffective.

Resistance of arm ACB, R₁ = R + R = 2R

Resistance of arm ADB, R₂ = R + R = 2R

Equivalent resistance between A and B, RAB is given by

$\frac{1}{\text{R}_\text{AB}}=\frac{1}{2\text{R}}+\frac{1}{2\text{R}}=\frac{2}{2\text{R}}$

$\Rightarrow\text{R}_\text{AB}=\text{R}=2\Omega$

2. In arm CD, there is no current, ICD = 0,

Current through arm ACB

$\text{i}_1=\frac{\text{V}}{\text{R}_1}$

$=\frac{10}{2\text{R}}=\frac{10}{2\times2}=\frac{10}{4}=2.5\text{A}$

1. $\text{At}^2=\text{Q}$

$\Rightarrow\text{A}=\frac{\text{Q}}{\text{t}^2}=\frac{\text{A}'\text{T}'}{\text{T}^{-2}}=\text{A}^1\text{T}^{-1}$

2. $\text{Bt}=\text{Q}$

$\Rightarrow\text{B}=\frac{\text{Q}}{\text{t}}=\frac{\text{A}'\text{T}'}{\text{T}}=\text{A}$

3. $\text{C}=[\text{Q}]$

$\Rightarrow\text{C}=\text{A}'\text{T}'$

4. Current $\text{t}=\frac{\text{dQ}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{At}^2+\text{Bt}+\text{C})$

$=2\text{At}+\text{B}=2\times5\times5+3=53\text{A}.$

1. $\text{at t}=0,\text{q}=60\mu\text{c}$

2. $\text{at t}=30\mu\text{s},\text{ q}=\text{Qe}^{\frac{-\text{t}}{\text{RC}}}$

$=60\times10^{-6}\times\text{e}^{-0.3}=44\mu\text{C}$

3. $\text{at t}=120\mu\text{s},\text{ q}=60\times10^{-6}\times\text{e}^{-1.2}=18\mu\text{c}$

4. $\text{at t}=1.0\text{ms, q}=60\times10^{-6}\times\text{e}^{-10}$ $=0.00272=0.003\mu\text{C}.$

1. $\text{R}=\frac{\text{V}^2}{\text{P}}$

$=\frac{250\times250}{500}=125\Omega$

2. $\text{A}=0.5\text{mm}^2$

$=0.5\times10^{-6}\text{m}^2$

$=5\times10^{-7}\text{m}^2$

$\text{R}=\frac{\text{fI}}{\text{A}}=\text{l}=\frac{\text{RA}}{\text{f}}$

$=\frac{125\times5\times10^{-7}}{1\times10^{-6}}$

$=625\times10^{-1}=62.5\text{m}$

3. $62.5=2\pi\text{r}\times\text{n}$

$62.5=3\times3.14\times4\times10^{-3}\times\text{n}$

$\Rightarrow\text{n}=\frac{62.5}{2\times3.14\times4\times10^3}$

$\Rightarrow\text{n}=\frac{62.5\times10^{-3}}{8\times3.14}\approx2500\ \text{turns}$

1. $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{24}=0.25\text{A}$

2. $\text{q}=\text{Q}\Big(1-\text{e}^{\frac{-\text{t}}{\text{RC}}}\Big)$

$=(8\times10^{-6}\times6)\big[1-\text{c}^{-1}\big]$

$=48\times10^{-6}\times0.63=3.024\times10^{-5}$

$\text{V}=\frac{\text{Q}}{\text{C}}=\frac{3.024\times10^{-5}}{8\times10^{-6}}=3.78$

$\text{E}=\text{V}+\text{iR}$

$\Rightarrow6=3.78+\text{i}24$

$\Rightarrow\text{i}=0.09\mathring{\text{A}}$

1. For minimum power consumed V₁ = 220 - 1% = 220 – 2.2 = 217.8

$\therefore\text{i}=\frac{\text{V}_1}{\text{R}}=\frac{217.8}{484}=0.45\text{A}$

Power consumed $=\text{i}\times\text{V}_1=0.45\times217.8=98.01\text{W}$

2. for maximum power consumed V₂ = 220 + 1% = 220 + 2.2 = 222.2

$\therefore\text{i}=\frac{\text{V}_2}{\text{R}}=\frac{222.2}{484}=0.459$

Power consumed = i × V₂ = 0.459 × 222.2 = 102W

1. No.of electrons transferred

$\text{i}=\frac{\text{V}}{\text{R}}=\frac{20}{10^3}=20\times10^{-3}=2\times10^{-2}\text{A}$

$\text{q}=\text{it}=2\times10^{-2}\times1=2\times10^{-2}\text{C}$

No. of electrons transferred $=\frac{2\times10^{-2}}{1.6\times10^{-19}}=\frac{2\times10^{-17}}{1.6}=1.25\times10^{17}.$

2. Current density of wire

​​​​​​​​​​​​​​​​​​​​​$=\frac{\text{i}}{\text{A}}=\frac{2\times10^{-2}}{\pi\times10^{-8}}=\frac{2}{3.14}\times10^6$

$=0.6369\times10^6=6.37\times10^5\text{A}/\text{m}^2.$

1. $\text{q}=\text{VC}\Big(1-\text{e}^{\frac{-\text{t}}{\text{CR}}}\Big)$

I = Current $=\frac{\text{dq}}{\text{dt}}=\text{VC}.(-)\text{e}^{\frac{-\text{t}}{\text{RC}}},\frac{-1}{\text{RC}}$

$=\frac{\text{V}}{\text{R}}\text{e}^{\frac{-\text{t}}{\text{RC}}}=\frac{\text{V}}{\text{R}\cdot\text{e}^{\frac{\text{t}}{\text{RC}}}}=\frac{24}{10^6}\times\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$

$=24\times10^{-6}\frac{1}{\text{e}^{\frac{\text{t}}{100}}}$

$\text{t}=10\text{min},600\text{sec}.$

$\text{Q}=24\times10+-4\times\big(1-\text{e}^{-6}\big)=23.99\times10^{-4}$

$\text{I}=\frac{24}{10^6}\times\frac{1}{\text{e}^6}=5.9\times10^{-8}\text{Amp}.$

2. $\text{q}=\text{VC}\Big(1-\text{e}^\frac{-\text{t}}{\text{CR}}\Big)$

1. For the loop PASQ $(\text{i}_1+\text{i}_2)\frac{15}{6}\text{rx}+\frac{15}{6}(6-\text{x})\text{i}_1+\text{i}_1\text{R}=\text{E}\ ...(1)$

For the loop AWTM, $-\text{i}_2.\text{R}-\frac{15}{6}\text{rx}(\text{i}_1+\text{i}_2)=\frac{\text{E}}{2}$

$\Rightarrow\text{i}_2\text{R}+\frac{15}{6}\text{r}\times(\text{i}_1+\text{i}_2)=\frac{\text{E}}{2}\ ...(2)$

For zero deflection galvanometer $\text{i}_2=0\Rightarrow\frac{15}{6}\text{rx}.\text{i}_1=\frac{\text{E}}{2}=\text{i}_1=\frac{\text{E}}{5\text{x}.\text{r}}$

Putting $\text{i}_1=\frac{\text{E}}{5\text{x . r}}$ and i₂ = 0 in equation (1), we get x = 320cm.

2. Putting x = 5.6 and solving equation (1) and (2) we get $\text{i}_2=\frac{3\text{E}}{22\text{r}}.$