Oswal Practice Papers CBSE Class 12 Physics Solutions (Practice Paper - 10)

Section-A

1. (b) increase

Explanation :

As the iron rod is removed from the coil, the coil’s self-inductance falls, the circuit’s impedance drops, and the current rises as a result. Therefore, the electricity used by the bulb grows, as does the brightness of the bulb.
Hence option (b) is correct.

2. (d) Both electric and magnetic field vectors are parallel to each other.

Explanation :

In electromagnetic waves,

The electric field and magnetic field vary continuously with time and have maxima and minima at the same place and at the same time.
Both electric and magnetic fields have the same energy.
Both electric and magnetic fields are perpendicular to each other and perpendicular to direction of propagation.
These waves don’t require any material medium to propagate, they can propagate in vacuum as well.

Hence option (d) is correct.

3. (b) the circular and elliptical loops

Explanation :

When the loops are being drawn out of the field, the rate at which the number of field lines cutting through the loop decreases will be a constant for square and rectangle loops, but not for circular or elliptical shapes.
Area coming out per second from the magnetic field is not constant for elliptical and circular loops, so induced emf, during the passage of these loops, out of the field region will not remain constant for the circular and the elliptical loops.
Hence option (b) is correct.

4. (d) 0.85

Explanation :

$$\text{Given path difference,} Δx =\frac{\lambda}{8}\\\text{Phase difference,}Δ\phi=\frac{2\pi}{\lambda}Δx\\Δ\phi=\frac{2\pi}{\lambda}\frac{\lambda}{8}\\Δ\phi=\frac{\pi}{4}\\I =I_0cos^2(\frac{Δ\phi}{2})\\\frac{I}{I_0}=cos^2(\frac{\pi}{8})\\\frac{I}{I_0}=0.85$$

In Young’s double slit experiment, the path difference at a certain point on the screen between two interfering waves is 1/8 th of the wavelength. The ratio of intensity at this point to that at the centre of a bright fringe is close to 0.85.
Hence option (d) is correct.

5. (b) will be in opposite direction

Explanation :

The direction of electric field at equatorial point A or B will be in opposite direction, as that of direction of dipole moment.

6. (d) 4.77 × 10^–3 m

Explanation :

The radius of the innermost orbit of a hydrogen atom, r₁ = 5.3 × 10−11 m. The radius of the orbit is proportional to the square of n.
Let r₂ be the radius of the orbit at n = 3.
r₃ = r₁ × 3²
= 5.3 × 10^–11 × 9
= 4.77 × 10–10 m
Hence, the radii of an electron for n = 3 orbits is 4.77 × 10⁻¹⁰ m.
Hence option (d) is correct.

7. (b) q₁ q₂ < 0

Explanation :

According to the Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges considered to
be point charges.

$$i.e., F=\frac{q_1q_2}{4\pi\epsilon_0r^2}$$

F₁₂ = − F₂₁
Here F₁₂ is the force exerted by q₁ on q₂ and F₂₁ is the force exerted by q₂ on q₁. There is a force of attraction and the charge must be unlike charge.
For attractive force, q₁ q₂ < 0
q₁ = + ve
q₂ = − ve
q₁q₂ < 0
The correct option is (b).

8. (d) Zero

Explanation :

The electric potential at a point due to an electric dipole is given by:
$$i.e.,V =\frac{kP cos\theta}{r^2}\\\text{For an equatorial plane,}\\\theta=\frac{\pi}{2}\\\text{and we know that cos 90° = 0.}\\\text{So, the potential at a point at the same distance on its equatorial line is given by,}\\V =\frac{kP cos90}{r^2}\\V = 0\\\text{Hence option (d) is correct.}$$

9. (b) material A is germanium and material B is copper

Explanation :

For semiconductors, there are very few free electrons. The conductor’s resistance decreases as a result of an increase in the number of free electrons that can act as carriers as a result of rising temperature [Fig. (i)]
For a conductor, when temperature rises, more free electrons become available. As a result, more of them begin to oppose one another’s motion, increasing the conductor’s resistance [Fig. (ii)]

We know germanium is a semiconductor while copper is a conductor. So, fig. (i) is a semiconductor while fig. (ii) is a conductor.
Hence, option (b) is correct.

10. (d) 6 A in the clockwise direction

Explanation :

Given : Radius (R₁) = 2x
Radius (R₂) = 3x
Current (I₂) = 9 A (anticlockwise)
To find : I₁
We know that the magnetic field at the common centre due to the loop is zero.
So we can say that magnetic field magnitude will be equal but direction is opposite.
Therefore,
$$\text{Therefore,}\space B_1=B_2\\B=\frac{\mu_0 I}{2R}\\\text{By using above relation,}\\\frac{\mu_0I_1}{2R_1}=\frac{\mu_0I_2}{2R_2}\\I_1=I_2×\frac{R_1}{R_2}\\I_1=9×\frac{2}{3}\\I_1=6A\\\text{But in opposite direction i.e., clockwise. Hence,option (d) is correct.}$$

11. (c) 4 : 3

Explanation :

$$\text{We know that magnetic field B in inside points =}\frac{\mu_0 Ir}{2\pi R^2}\\\text{(where R is the radius and r = distance of the inside point from the axis and i = total current)}\\\text{Distance from the axis at}\frac{a}{2}\text{inside =}(a-\frac{a}{2})=\frac{a}{2}\\\text{So magnetic field at}\frac{a}{2}\text{inside =}\frac{\mu_0×i×a}{2×(2\pi a^2)}=\frac{\mu_0 i}{4\pi a}\\\text{Distance from the axis at}\frac{a}{2}\text{outside =}\frac{a+a}{2}=\frac{3a}{2}\\\text{We know that magnetic field B in outside points =}\frac{\mu_0 i}{2\pi r}\\\text{(r = distance of the outside point from the axis and i = total current)}\\\text{So magnetic field at}\frac{a}{2}\text{outside =}\frac{\mu_0×i}{(3a/2)×2\pi}\\=\frac{\mu_0 i}{3\pi a}\\\frac{\text{Magnetic field at outside}}{\text{Magnetic field at inside}}=(\frac{\mu_0 I×4\pi a}{\mu_0 I×3\pi a})=frac{4}{3}$$

Hence option (c) is correct.

12. (a) decreases

Explanation :

The magnetic permeability of a substance is defined as,
μ = B/H
where B is the established magnetic field inside the material and H is the applied external magnetic field.
When the external applied magnetic field H is increased, the established magnetic field for a ferromagnetic material remains the same.
Thus, the magnetic permeability decreases.
Hence option (a) is correct.

13. (c) A is true but R is false.

Explanation :

This is true that electrical conductivity of a semiconductor increases on doping. In semiconductors, doping can be done with either electron rich or electron deficient impurities.
Addition of such impurities cause the electronic defects in a crystal structure, which enhances the electrical conductivity. So, reason is wrong as doping always does not increase, the number of electrons in the semiconductor.
Hence. option (c) is correct.

14. (c) A is true but R is false.

Explanation :

We know that fringe width is proportional to D/d. In an interference pattern observed in Young’s double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to (1/3)rd, then new fringe width
remains the same. This is true.
$$\text{Fringe width}=\frac{\frac{1}{3} D}{\frac{1}{3}d}\space\text{Fringe width=Remains same.}\\\text{Hence option (c) is correct.}$$

15. (a) Both A and R are true and R is the correct explanation of A.

Explanation :

Different photoelectrons emit electrons at various speeds. Actually, not all electrons occupy the same energy level; instead, they do so in a continuous range of levels. As a result, the energy of the electrons that are knocked off at different levels vary. The energy of electrons emitted from inside the metal surface is lost in collision with the other atoms in the metal.
Hence, option (a) is correct.

16. ((a) A is true but R is false.

Explanation :

series resonance circuit, inductive reactance is equal to capacitive reactance.
$$i.e., \omega L=\frac{1}{\omega C}\\Z=\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}\\=R$$

Section-B

17. (i) λ₁ → Microwaves are considered suitable for radar systems used in aircraft navigation because they have a short wavelength range (10–3 m to 0.3 m), which makes them suitable for long range communication.

(ii) λ₂ → Infrared rays, Infrared is used to improve visibility on runways during fog and mist conditions. So, it is the wavelength of infrared waves.

Ascending order of wavelength in ascending order:
Wavelength of UV rays < Wavelength of Infrared < wavelength of microwave i.e., λ₂ < λ₃ < λ₁

18. (i) A is diamagnetic and B is ferromagnetic.
(ii) Diamagnetic materials have permeabilities less than 1 (one) and have negative susceptibility. Their atoms and molecules do not have permanent dipole moment. The field lines get expelled in them. Ferromagnetic materials have permeability more than one and susceptibility positive. Their atoms and molecules have permanent dipole moment. So the field lines get concentrated in them.

19. Generally, a biconvex lens acts as a converging lens in air because the refractive index of air is less than that of the material of the lens. The refractive index of water (1.33) is more than the refractive index of the material of the lens (1.25). So, on immersing in water it will behave as a diverging lens.
$$\frac{1}{f}=(\mu-1)(\frac{1}{R_1}-\frac{1}{R_2})\\\frac{1}{f}=(\frac{\mu_m}{\mu_\omega}-1)(\frac{1}{R_1}-\frac{1}{R_2})\\\frac{\mu_m}{\mu_\omega}=\frac{1.25}{1.33}\\\frac{\mu_m}{\mu_\omega}=0.98$$

Where f is the focal length of the lens, µm is the refractive index of material of lens and µw is the refractive index of water.
The value of (µ – 1) is negative and focal length ‘f ’ will be negative. So, it will behave like a diverging lens.

20. Given: λ = 6000 Å
We know,
Angular width 2θ =2λ/d
In case of new λ (assumed λ′ here), angular width decreases by 30%
$$\frac{2\lambda'}{d}=0.70×2(\frac{2\lambda}{d})\\\frac{2\lambda'}{d}=\frac{(0.70×2×6000)}{d}\\λ′=\frac{8400}{2}\\λ′=4200 Å$$

Intensity pattern for double slit interference :

21. As the semiconductors at room temperature act as insulators so they offer a very high resistance value. But with the increase in temperature, they start conducting by decreasing the energy-gap and hence they offer less resistivity. So, the resistivity of semiconductors decreases with the increase in temperature and as a result resistance also decreases. So to maintain the constant reading on the ammeter, we should increase the value of variable resistance R to compensate with the decreased value of resistance for the semiconductor.
Hence, to keep the reading of ammeter constant value of R should be increased as with the increase in temperature of a semiconductor, its resistance decreases and current tends to increase.

In case of reverse biased diodes, the potential barrier becomes higher as the battery further raises the potential of the n-side. The effective barrier height under reverse biased is (V₀ + V).
C-forward biased
Due to forward bias connection, the potential of the p side is raised and hence the height of the potential barrier decreases therefore, the effective barrier height is reduced to (V₀ – V).

Section-C

22. Figure shows an AC source, generating a voltage e = e0 sin wt, connected to a capacitor of capacitance C. The plates of the capacitor get charged due to the applied voltage. As the alternating voltage is reversed in each half cycle,

An AC source connected to a capacitor

the capacitor is alternatively charged and discharged. If q is the charge on the capacitor, the corresponding potential difference across the plates of the capacitor is V =q/C
q = CV.
q and V are functions of time, with V = e = e₀ sin wt. The instantaneous current in the circuit
$$\text{is i}=\frac{dq}{dt}=\frac{d}{dt}(CV)=C\frac{dV}{dt}\\= C\frac{d}{dt}(e_0sin\omega t)\\=\omega Ce_0 cos\omega t\\i =\frac{e_0}{1/\omega C}sin(\omega t+\frac{\pi}{2})\\= i_0 sin(\omega t+\frac{\pi}{2})\\Where i_0=(\frac{e_0}{1/\omega C})\text{is the peak value of the current.}$$

ωt (rad)	Mωt + π/2 (rad)	e = e₀ sin ωt	i = i₀ sin(ωt+π/2)
0	π/2	0	i₀
π/2	π	E₀	0
π	3π/2	0	-i₀
3π/2	2π	-e₀	0
2π	2π+π/2	0	-i₀

Table gives the values of e and i for different values of cot and Fig. shows graphs of e and i versus ωt. i leads e by a phase angle of π/2 rad.

AC circuit containing pure inductance: Consider a coil of self-inductance L and negligible ohmic resistance. An alternating potential difference is applied across its ends. The magnitude and direction of AC changes periodically, due to which there is a continual change in magnetic flux linked with the coil. Therefore according to Faraday’s law, an induced emf is produced in the coil, which opposes the applied voltage. As a result the current in the circuit is reduced. That is inductance acts like a resistance in ac circuit. The instantaneous value of alternating voltage applied

V = V₀ sin ωt.

If i is the instantaneous current in the circuit and di/dt the rate of change of current in the circuit at that instant, then instantaneous induced emf
$$e\epsilon=−L\frac{di}{dt}\\\text{According to Kirchhoff’s loop rule}\\V+\epsilon=0⇒V−L\frac{di}{dt}=0\\or\frac{di}{dt}=\frac{V}{L}\\or\frac{di}{dt}= V_0 sin ωt/L\\or\space di=V_0 sin ωt/L.dt\\\text{Integrating with respect to time ‘t’:}\\i=\frac{V_0}{\omega L}\int sin \omega tdt=\frac{V_0}{L}(-\frac{cos\omega t}{\omega})\\=-\frac{V_0}{\omega L} cos \omega t=-\frac{V_0}{\omega L}sin(\frac{\pi}{2}-\omega t)\\=\frac{V_0}{\omega L}sin(\omega t-\frac{\pi}{2})...(ii)\\\text{This is required expression for current}\\or i=i_0 sin e(\omega t-\frac{\pi}{2})\\and i_0=\frac{V_0}{\omega L}...(iv)$$

Where i₀ is the peak value of alternating current.
Also comparing (i) and (iii), we note that current lags behind the applied voltage by an angle 2π .

23. Consider a small length L of the long straight conductor
$$B_1=\frac{\mu_0I_1}{2\pi d}\\B_2=\frac{\mu_0I_2}{2\pi d}\\F_{12}=I_2B_1L\\F_{21}=I_1B_2L\\\text{The total force on conductor of length L will be}\\F=F_{12}=F_{21}=\frac{\mu_0I_2I_2L}{2\pi d}\\\text{Force acting per unit length of conductor}\\f=\frac{F}{L}=\frac{\mu_0I_2I_2L}{2\pi d}$$

Ampere is that value of steady-state current, which when maintained in each of the very two long,
straight, parallel conductors of negligible cross-section and placed 1m apart in a vacuum produces
2 × 10^–7 N/m force on each conductor.

24. Area of the circular loop = πr²

$$= 3.14×(0.12)^2m^2\\=4.5×10^{–2}m^2\\E=–\frac{\phi}{dt}\\=\frac{-d}{dt}(BA)=-A\frac{dB}{dt}\\E = -A(\frac{B_2-B_1}{t_2-t_1})\\For 0 < t < 2s\\E_1 = – 4.5 × 10^{–2} ×(\frac{(1-0)}{(2-0)})\\= –2.25 × 10^{–2} V\\I_1=\frac{E_1}{R}=-\frac{-2.25× 10^{–2}}{8.5}\\A = – 2.6 × 10^{–3} A\\I_1 = –2.6 mA\\For 2s < t < 4s,\\E_2 = – 4.5 × 10^{–2} ×(\frac{(1-1)}{(4-2)})=0\\I_2 =\frac{E_2}{R}=0\\For 4s < t < 6s,\\I_3 =-\frac{(4.5 × 10^{–2})}{8.5}×\frac{(0-1)}{6-4}= 2.6 mA$$


	0 < t < 2s	2 < t < 4s	4 < t < 6s
E(V)	– 0.023	0	+ 0.023
I(mA)	– 2.6	0	+2.6

25.

	n-type semiconductors	p-type semiconductors
(i)	It is an extrinsic semiconductor obtained by adding a pentavalent impurity to a pure intrinsic semiconductor.	It is also an extrinsic semiconductor obtained by adding a trivalent impurity to a pure intrinsic semiconductor.
(ii)	The impurity atoms added provide extra free electrons to the crystal lattice and are called donor atoms.	The impurity atoms added create holes in the crystal lattice and are called acceptor atoms.
(iii)	The electrons are majority carriers and the holes are minority carriers.	The holes are majority carriers and the electrons are minority carriers.
(iv)	The electrons concentration is much more than the hole concentration (n_e >> n_h).	The hole concentration is much more than the electron concentration (n_h >> n_e).

26. (a) A – cut off or stopping potential
X – anode potential

Variation of photoelectric current with collector plate potential for different frequencies of incident radiation

Variation of photocurrent with collector plate potential for different intensity of incident radiation

27. A transition from n = 3 to n = 1 state, the energy of the emitted photon
$$h_v=E_2−E_1\\=13.6[\frac{1}{1^2}-\frac{1}{3^2}]eV\\=12.1eV\\\text{From Einstein’s photoelectric equation,}\\hv=K_{max}+W_0\\W_0=hv−K_{max}\\=12.1−9\\=3.1eV\\\text{Threshold wavelength,}\\λ_0=\frac{hc}{W_0}\\=6.62×10^{−34}×3×\frac{10^8}{3.1×1.6×19^{-19}}\\=4×10^{−7}m$$

28. Since, the electric field is parallel to the faces parallel to xy and xz planes, the electric flux through them is zero.

$$\text{Electric flux through the left face}\\φL=(E_L)(a^2)cos180°\\=(0) (a^2)cos180°=0\\\text{Electric flux through the right face,}\\φR=(E_R)(a^2)cos0°\\=(2a)(a^2)×1=2a^3\\\text{Total flux(φ)}= 2a^3=\frac{q_{enclosed}}{\epsilon_0}\\q_{enclosed}=2a^3ε_0$$

(i) Net force on the charge Q/2, placed at the centre of the shell is zero.
Force on charge 2Q kept at a point A
$$F=E×2Q=\frac{1.(\frac{3Q}{2})2Q}{4\pi\epsilon_0r^2}=\frac{k×3Q^2}{r^2}\\where\space k=\frac{1}{4\pi\epsilon_0r^2}\\\text{(ii) Electric flux through the shell,}\\\phi =\frac{Q}{2\epsilon_0}$$

Section-D

29.
(i) (d) microwaves

(a) Hertz
(ii) (c) velocity
(iii) (c) wavelength is halved and frequency remains unchanged.
(iv) (a) u_e = u_m

30.
(i) (c) Eg > 3 eV
(ii) (b) 1 eV
(iii) (a) conductors
(iv) (a) The number of free electrons for conduction is significant only in Si and Ge but small in C.

(b) conductor

Section-E

31. (a)

$Diffraction angle$

(b)

$$\text{If iC is the critical angle for the prism/material,}\\\mu=\frac{1}{sini_C}\\sin i_C=\frac{1}{\mu}=\frac{\sqrt{3}}{2}\\⇒i_C=60°\\\text{Angle of incidence at face AC of the prism=60°}\\\text{Hence, refracted ray grazes the surface AC.}\\⇒Angle of emergence=90°\\⇒Angle of deviation=30°$$

(a) (i) The interference pattern has a number of equally spaced bright and dark bands. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima. The intensity falls as we go to successive maxima away from the centre, on either side.

(ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

$$(b)(i)\space\mu=\frac{sin(\frac{A+\delta_m}{2})}{sin(\frac{A}{2})}=\frac{sin(\frac{60+30}{2})}{sin(\frac{60}{2})}=\sqrt{2}\\Also\space\mu=\frac{c}{v}\\⇒v=\frac{3×10^8}{\sqrt{2}}m/s$$

(ii) At face AC, let the angle of incidence be r₂. For grazing ray,

$$e = 90°\\⇒ \mu =\frac{1}{sin r_2}\\⇒ r_2 = sin^{-1}(\frac{1}{\sqrt{2}})=45\\\text{Let angle of refraction at face AB be }r_1.\\Now\space r_1 + r_2=A\\r_1=A–r_2=60°–45°=15°\\\text{Let angle of incidence at this face be i}\\\mu=\frac{sin i}{sin r_1}\\⇒\sqrt{2}=\frac{sin i}{sin 15}\\i = sin^{-1}(\sqrt{2}.sin 15)=215$$

32. (a) (i) a dipole

(ii) two identical positive charges.

(b) In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^–3 m² and the separation between the plates is 3 mm.
$$Here,A=6×10^{–3}m^2,d=3mm=3×10^{–3}m\\\text{(i) Capacitance,}\\C=\frac{\epsilon_0A}{d}=(\frac{8.85×10^{-12}×6×10^{-3}}{3×10^{-3}})=17.7×10^{-12}F\\(ii) Charge, Q=CV=17.7×10^{–12}×100=17.7×10^{–10}C\\(iii) New charge Q’=KQ=6×17.7×10^{–10}=1.062×10^{–8}C$$

(a) We know potential energy is a scalar quantity
$$\text{Now total P.E.}=\frac{k(-q)Q}{x}+\frac{k(-q)Q}{x}+\frac{k(-q)(-q)}{2x}\\\text{Given potential energy = 0}\\0=\frac{-2kqQ}{x}+\frac{kq^2}{2x}\\0=–4Q+q\\Q:q=1:4$$

(b) Consider a thin spherical shell of radius R with a positive charge q distributed uniformly on the surface. As the charge is uniformly distributed, the electric field is symmetrical and directed radially outward.

(i) When point P lies outside the spherical shell: Suppose that we have calculated the field at the point P at a distance r(r > R) from its centre. Draw Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical surface of radius r and centre O.
Let E be the electric field at point P, then the electric flux through area element of area ds is given by
dΦ = E.ds
Since ds is also normal to the surface.
Total electric flux through the Gaussian surface is given by
$$\phi=∫Eds=E∫ds\\Now, ∫ds=4\pi r^2 ...(i)\\=E×4\pi r^2\\\text{Since the charge enclosed by the Gaussian surface is q, according to the Gauss’s theorem,}\\\phi =\frac{q}{\epsilon_0}...(ii)\\\text{From equation (i) and (ii) we obtain}\\E×4\pi r^2=\frac{q}{\epsilon_0}\\E=\frac{q}{4\pi r^2\epsilon_0}(for r>R)$$

(ii) The graph showing the variation of the electric field as a function of r is shown below

33. (a) Drift velocity: It is the average velocity acquired by the free electrons superimposed over the random motion in the direction opposite to electric field and along the length of the metallic conductor.
If e is the magnitude of charge on each electron, then the total charge in the conductor,
Q = (nAl)_e ...(i)
The time taken by the charge to cross the conductor length is given by
$$t=\frac{l}{v_d}$$

where v_d is the drift velocity of electrons.
According to the definition of electric current,
$$I=\frac{Q}{t}=\frac{nAle}{\frac{l}{v_d}}\\I = neAv_d\\I = I_1 + I_2 ...(i)\\(b) Here,...(i)\\\text{Let V = Potential difference between A and B. For cell ε1 Then,}\\V = \epsilon_1 – I_1 r_1\\⇒ I_1 =\frac{\epsilon_1-V}{r_1}\\\text{Similarly, for cell}\epsilon_2,\\I_2=\frac{\epsilon_2-V}{r_2}\\\text{Putting these values in equation (i),}\\I =\frac{\epsilon_1-V}{r_1}+\frac{\epsilon_2-V}{r_2}\\or I =(\frac{\epsilon_1}{r_1}+\frac{\epsilon_2}{r_2})-V(\frac{1}{r_1}+\frac{1}{r_2})$$

Comparing the above equation with the equivalent circuit of emf ‘ε_eq’ and internal resistance ‘r_eq’ then,

$$Then V=ε_{eq}–Ir_{eq}...(iii)\\(i)ε_{eq}=\frac{ε_1r_2+ε_2r_2}{r_1+r_2}...(iii)\\(ii)r_{eq}=\frac{r_1r_2}{r_1+r_2}\\\text{(iii) The potential difference between A and B}\\V=ε_{eq}–Ir_{eq}$$

(a) (i) Kirchhoff’s first law (junction rule) : In an electrical circuit, the algebraic sum of currents meeting at a junction is always zero.
(i_i) Kirchhoff’s second rule (loop rule) : In any closed mesh of an electrical circuit, the algebraic sum of the e.m.fs. is equal to the algebraic sum of products of resistances and current flowing
through them.

(b) Wheatstone bridge (balanced) : Let i be the current from battery E. At point A, current i1 flows through resistance P and current i − i₁ flows through R.
In balanced state, no current flows through BD, hence point B and D are at same potential.
Therefore, current i₁ flows through resistance Q also and current i − i₁ flows through S. Applying Kirchhoff’s loop rule in closed mesh ABDA,
$$i_1P−(i−i_1)R=0\space or\space i_1P=(i − i_1)R ...(i)\\\text{In closed mesh BCDB,}\\i_1Q−(i−i_1)S=0\space or\space i_1Q=(i − i_1)S ...(ii)\\\text{Dividing (i) from (ii), we get}\\\frac{P}{Q}=\frac{R}{S}\\\text{This is the condition for balance in a Wheatstone’s bridge}$$