43 + Interview Questions in GENERAL QA IN CAPACITANCE Page 1 InterviewSolution

1.	A multiple plate capacitor has 10 plates, each of area 10 square cm and separation between 2 plates is 1 mm with air as dielectric. Determine the energy stored when voltage of 100 volts is applied across the capacitor.
Answer» Number of plates, n = 10 C = ((n - 1)∈₀)/d = (9 x 8.854 x 10^-12 x 10 x 10^-4)/(1 x 10^-3) = 79.7pF Energy stored = 1/2 × 79.7 × 10^{− 12} × 100 × 100 = 0.3985 μJ

1.

A multiple plate capacitor has 10 plates, each of area 10 square cm and separation between 2 plates is 1 mm with air as dielectric. Determine the energy stored when voltage of 100 volts is applied across the capacitor.

Answer»

Number of plates, n = 10

C = ((n - 1)∈₀)/d = (9 x 8.854 x 10^-12 x 10 x 10^-4)/(1 x 10^-3) = 79.7pF

Energy stored = 1/2 × 79.7 × 10^{− 12} × 100 × 100 = 0.3985 μJ

Discussion

2.	Three capacitors of 2 μ F, 5 μ F and 10 μ F have breakdown voltage of 200 V, 500 V and 100 V respectively. The capacitors are connected in series and the applied direct voltage to the circuit is gradually increased. Which capacitor will breakdown first ? Determine the total applied voltage and total energy stored at the point of breakdown.
Answer» C₁ of 2μF, C₂of 5μ F, and C₃ of 10μF are connected in series. If the equivalent single capacitor is C, 1/C = 1/C₁ + 1/C₂ + 1/C₃, which gives C = 1.25 μ F If V is the applied voltage, V₁ = V × C/C₁ = V × (1.25/2) = 62.5 % of V V₂ = V × (C/C₂) = C × (1.25/5) = 25 % of V V₃ = V × (C/C₃) = V × (1.25/10) = 12.5 % of V If V₁= 200 volts, V = 320 volts and V₂ = 80 volts, V₃ = 40 volts. It means that, first capacitor C₁ will breakdown first. Energy stored = 1/2 CV₂ = 1/2 × 1.25 × 10^{− 6} × 320 × 320 = 0.064 Joule

2.

Three capacitors of 2 μ F, 5 μ F and 10 μ F have breakdown voltage of 200 V, 500 V and 100 V respectively. The capacitors are connected in series and the applied direct voltage to the circuit is gradually increased. Which capacitor will breakdown first ? Determine the total applied voltage and total energy stored at the point of breakdown.

Answer»

C₁ of 2μF, C₂of 5μ F, and C₃ of 10μF are connected in series.

If the equivalent single capacitor is C,

1/C = 1/C₁ + 1/C₂ + 1/C₃, which gives C = 1.25 μ F

If V is the applied voltage,

V₁ = V × C/C₁ = V × (1.25/2)

= 62.5 % of V

V₂ = V × (C/C₂) = C × (1.25/5) = 25 % of V

V₃ = V × (C/C₃) = V × (1.25/10) = 12.5 % of V

If V₁= 200 volts, V = 320 volts and V₂ = 80 volts, V₃ = 40 volts.

It means that, first capacitor C₁ will breakdown first.

Energy stored = 1/2 CV₂ = 1/2 × 1.25 × 10^{− 6} × 320 × 320 = 0.064 Joule

Discussion

3.	Device used to store energy in electrical circuits is A. resistor B. inductor C. capacitor D. diode
Answer» C. capacitor.

Discussion

4.	The voltage across a 5microF capacitor change uniformly from 10 to 70V in 5ms. Calculate the change in capacitor charge and charging current.
Answer» Given c = 5μF t = 5ms ΔV = 10 to 70V change in capacitor charge Δq = cΔV = 5 x 10^-6 (V_f- V_i) = 5 x 10^-6 (70 - 10) = 300 x 10^-6 Δq = 0.3 mC change in current i = c dv/dt i = 5 x 10^-6 x \(\frac{60}{5\times10^{-3}}\) i = 60 x 10^-3 i = 60 mA

4.

The voltage across a 5microF capacitor change uniformly from 10 to 70V in 5ms. Calculate the change in capacitor charge and charging current.

Answer»

Given

c = 5μF

t = 5ms

ΔV = 10 to 70V

change in capacitor charge Δq = cΔV

= 5 x 10^-6 (V_f- V_i)

= 5 x 10^-6 (70 - 10)

= 300 x 10^-6

Δq = 0.3 mC

change in current i = c dv/dt

i = 5 x 10^-6 x \(\frac{60}{5\times10^{-3}}\)

i = 60 x 10^-3

i = 60 mA

Discussion

5.	A resistance R and a 4 μF capacitor are connected in series across a 200 V. d.c. supply. Across the capacitor is a neon lamp that strikes (glows) at 120 V. Calculate the value of R to make the lamp strike (glow) 5 seconds after the switch has been closed.
Answer» Obviously, the capacitor voltage has to rise 120 V in 5 seconds. ∴ 120 = 200 (1 −e^{− 5/λ}) or e 5/λ = 2.5 or λ = 5.464 second. Now, λ = CR ∴ R = 5.464/4 × 10⁻⁶ = 1.366 MΩ

5.

A resistance R and a 4 μF capacitor are connected in series across a 200 V. d.c. supply. Across the capacitor is a neon lamp that strikes (glows) at 120 V. Calculate the value of R to make the lamp strike (glow) 5 seconds after the switch has been closed.

Answer»

Obviously, the capacitor voltage has to rise 120 V in 5 seconds.

∴ 120 = 200 (1 −e^{− 5/λ})

or e 5/λ = 2.5 or λ = 5.464 second.

Now, λ = CR

∴ R = 5.464/4 × 10⁻⁶ = 1.366 MΩ

Discussion

6.	A parallel-plate capacitor is made of plates 1 m square and has a separation of 1 mm. The space between the plates is filled with dielectric of εr = 25.0. If 1 k V potential difference is applied to the plates, find the force squeezing the plates together
Answer» F = − (1/2)ε₀ε_rAE² newton Now E = V/d = 1000/1 × 10⁻³ = 10⁻⁶ V/m ∴ F = - 1/2ε₀ε_r AE² = 1/2 x 8.854 x 10^-12 x 25 x 1 x (10⁶)² = - 1.1 x 10^-4N

6.

A parallel-plate capacitor is made of plates 1 m square and has a separation of 1 mm. The space between the plates is filled with dielectric of εr = 25.0. If 1 k V potential difference is applied to the plates, find the force squeezing the plates together

Answer»

F = − (1/2)ε₀ε_rAE² newton

Now E = V/d = 1000/1 × 10⁻³ = 10⁻⁶ V/m

∴ F = - 1/2ε₀ε_r

AE² = 1/2 x 8.854 x 10^-12 x 25 x 1 x (10⁶)² = - 1.1 x 10^-4N

Discussion

7.	Work done in charging a capacitor is given by A. (1⁄2)QV B. 2QV C. QV D. 2V
Answer» The Correct option is A. (1⁄2)QV.

Discussion

8.	Energy stored in a 2000 mF capacitor charged to a potential difference of 10 V is A. 0.12 J B. 1.3 J C. 0.10 J D. 3 J
Answer» The Correct option is C. 0.10 J

Discussion

9.	Capacitor is fully charged if potential difference is equal to A. e.m.f B. current C. resistance D. power
Answer» The Correct option is A. e.m.f

Discussion

10.	Total capacitance of 300 mF capacitor and a 600 mF in series isA. 300 mF B. 500 mF C. 200 mF D. 1000 mF
Answer» The Correct option is C. 200 mF

Discussion

11.	A single-core lead-sheathed cable, with a conductor diameter of 2 cm is designed to withstand 66 kV. The dielectric consists of two layers A and B having relative permittivities of 3.5 and 3 respectively. The corresponding maximum permissible electrostatic stresses are 72 and 60 kV/cm. Find the thicknesses of the two layers
Answer» g_max1/g_max2 = (ε_r2 x r₂)/(ε_r1 x r₁) or 72/60= (3 x r₂)/(3.5 x 1) or r₂ = 1.4cm Now, g_max = (V₁ x √2)/(2.3r₁log₁₀(r₂/r₁)) where V₁is the r.m.s. values of the voltage across the first dielectric. ∴ 72 = (V₁ x √2)/(2.3 x 1 x log₁₀1.4) or V₁ = 17.1kv Obviously, V₂= 60 − 17.1= 48.9 kV Now, g_max2 = (V₂x √2)/(2.3r₂log₁₀(r₃/r₂)) ∴ 60 = 48.9/(2.3 x 1.4log₁₀(r₃/r₂)) ∴ log₁₀ (r₃/r₂) = 0.2531 = log₁₀ (1.79) ∴ r₃/r₂ = 1.79 or r₃= 2.5cm Thickness of first dielectric layer = 1.4 − 1.0 = 0.4 cm. Thickness of second layer = 2.5 − 1.4 = 1.1 cm.

11.

A single-core lead-sheathed cable, with a conductor diameter of 2 cm is designed to withstand 66 kV. The dielectric consists of two layers A and B having relative permittivities of 3.5 and 3 respectively. The corresponding maximum permissible electrostatic stresses are 72 and 60 kV/cm. Find the thicknesses of the two layers

Answer»

g_max1/g_max2 = (ε_r2 x r₂)/(ε_r1 x r₁) or 72/60= (3 x r₂)/(3.5 x 1) or r₂ = 1.4cm

Now, g_max = (V₁ x √2)/(2.3r₁log₁₀(r₂/r₁))

where V₁is the r.m.s. values of the voltage across the first dielectric.

∴ 72 = (V₁ x √2)/(2.3 x 1 x log₁₀1.4) or V₁ = 17.1kv

Obviously, V₂= 60 − 17.1= 48.9 kV

Now, g_max2 = (V₂x √2)/(2.3r₂log₁₀(r₃/r₂))

∴ 60 = 48.9/(2.3 x 1.4log₁₀(r₃/r₂))

∴ log₁₀ (r₃/r₂) = 0.2531 = log₁₀ (1.79)

∴ r₃/r₂ = 1.79 or r₃= 2.5cm

Thickness of first dielectric layer = 1.4 − 1.0 = 0.4 cm.

Thickness of second layer = 2.5 − 1.4 = 1.1 cm.

Discussion

12.	In Hall effect, voltage across probe is known as A. hall voltage B. e.m.f C. potential difference D. hall potential
Answer» A. hall voltage.

Discussion

13.	Electric field strength related to hall voltage is given by A. VHd B. VH⁄d C. VHE D. Ed
Answer» The Correct option is B. V_H⁄_d

Discussion

14.	Force on a moving charge in a uniform magnetic field depends uponA. magnetic flux density B. the charge on the particle C. the speed of particle D. all of above
Answer» D. all of above

Discussion

15.	Hall voltage is directly proportional to A. current B. electric field C. magnetic flux density D. all of above
Answer» C. magnetic flux density

Discussion

16.	Force due to magnetic field and velocity is A. at right angles to each other B. at acute angles with each other C. at obtuse angle with each other D. antiparallel to each other
Answer» A. at right angles to each other

Discussion

17.	An electron is travelling at right angles to a uniform magnetic field of flux density 1.2 mT with a speed of 8 × 106 m s-1, the radius of circular path followed by electron isA. 3.8 cm B. 3.7 cm C. 3.6 cm D. 3.5 cm
Answer» The Correct option is A. 3.8 cm

Discussion

18.	Ahorizontalplane-platecapacitorischargedtoapotentialdifferenceof3,000V.Theseparationdistanceoftheplatesis10cm.Amercurydropcarryingacharge10-6Cmovesupinsidethecapacitorwithacceleration2.2m/s2.Determinethemassofthedrop.Whatistheweightofthedropinsidethecapacitor?
Answer» As given data is V = 3000 v , d = 10 cm = 0.1 m ,q = 10^(-6) c & a = 2.2 m/s^2.... Now, As we know F = qE & V = Ed ; ma = qE i.e, m2.2 = 10^(-6) (V/d) So, m = 0.0136 kg i.e, 13.6 g. Therefore, Weight = m*g = 0.136 ...

Discussion

19.	A capacitor of 10 pF is connected to a voltage source of 100 V. If the distance between the capacitor plates is reduced to 50 % while it remains, connected to the 100 V supply. Find the new values of charge, energy stored and potential as well as potential gradient. Which of these quantities increased by reducing the distance and why ?
Answer» (i) C = 10 pF (ii) C = 20 pF, distance halved Charge = 1000 p Coul Charge = 2000 p-coul Energy = 1/2 CV² = 0.05 μ J Energy = 0.10 μ J Potential gradient in the second case will be twice of earlier value

19.

A capacitor of 10 pF is connected to a voltage source of 100 V. If the distance between the capacitor plates is reduced to 50 % while it remains, connected to the 100 V supply. Find the new values of charge, energy stored and potential as well as potential gradient. Which of these quantities increased by reducing the distance and why ?

Answer»

(i) C = 10 pF

(ii) C = 20 pF, distance halved

Charge = 1000 p Coul

Charge = 2000 p-coul

Energy = 1/2 CV² = 0.05 μ J

Energy = 0.10 μ J

Potential gradient in the second case will be twice of earlier value

Discussion

20.	A liquid resistor consists of two concentric metal cylinders of diameters D = 35 cm and d = 20 cm respectively with water of specific resistance ρ = 8000 Ω cm between them. The length of both cylinders is 60 cm. Calculate the resistance of the liquid resistor.
Answer» r₁ = 10 cm ; r₂ = 17.5 cm; log₁₀ (1.75) = 0.243 ρ = 8 × 10³ Ω − cm; l = 60 cm. Resistance of the liquid resistor R = (2.3 x 8 x 10³)/(2π x 60) x 0.243 = 11.85Ω

20.

A liquid resistor consists of two concentric metal cylinders of diameters D = 35 cm and d = 20 cm respectively with water of specific resistance ρ = 8000 Ω cm between them. The length of both cylinders is 60 cm. Calculate the resistance of the liquid resistor.

Answer»

r₁ = 10 cm ; r₂ = 17.5 cm;

log₁₀ (1.75) = 0.243

ρ = 8 × 10³ Ω − cm; l = 60 cm.

Resistance of the liquid resistor

R = (2.3 x 8 x 10³)/(2π x 60) x 0.243 = 11.85Ω

Discussion

21.	The capacitance of a capacitor formed by two parallel metal plates each 200 cm2 in area separated by a dielectric 4 mm thick is 0.0004 micro farads. A p.d. of 20,000 V is applied. Calculate (a) the total charge on the plates (b) the potential gradient in V/m (c) relative permittivity of the dielectric (d) the electric flux density.
Answer» C = 4 × 10⁻⁴μF ; V = 2 × 10⁴ V (a) ∴ Total charge Q = CV = 4 × 10⁻⁴ × 2 × 10⁴ μC = 8 μC = 8 × 10⁻⁶ C (b) Potential gradient = dv/dx = (2 x 10⁴)/(4 x 10^-3) = 5 x 10⁶V/m (c) D = Q/A = 8 × 10⁻⁶ /200 × 10⁻⁴ = 4 × 10⁻⁴ C/m² (d) E = 5 × 10⁶ V/m Since D = ε₀ ε_r E ∴ ε_r = D/(ε₀ x E) = (4 x 10^-4)/(8.854 x 10^-12 x 5 x 10⁶) = 9

21.

The capacitance of a capacitor formed by two parallel metal plates each 200 cm2 in area separated by a dielectric 4 mm thick is 0.0004 micro farads. A p.d. of 20,000 V is applied. Calculate (a) the total charge on the plates (b) the potential gradient in V/m (c) relative permittivity of the dielectric (d) the electric flux density.

Answer»

C = 4 × 10⁻⁴μF ; V = 2 × 10⁴ V

(a) ∴ Total charge Q = CV = 4 × 10⁻⁴ × 2 × 10⁴ μC = 8 μC = 8 × 10⁻⁶ C

(b) Potential gradient = dv/dx = (2 x 10⁴)/(4 x 10^-3) = 5 x 10⁶V/m

(c) D = Q/A = 8 × 10⁻⁶ /200 × 10⁻⁴ = 4 × 10⁻⁴ C/m²

(d) E = 5 × 10⁶ V/m

Since D = ε₀ ε_r E

∴ ε_r = D/(ε₀ x E) = (4 x 10^-4)/(8.854 x 10^-12 x 5 x 10⁶) = 9

Discussion

22.	Calculate the values of i2, i3, v2, v3, va, vc and vL of the network shown in Fig. 5.40 at the following times : (i) At time, t = 0 + immediately after the switch S is closed ; (ii) At time, t → ∞ i.e. in the steady state.
Answer» (i) In this case the coil acts as an open circuit, hence i₂ = 0 ; v₂ = 0 and v_L = 20 V. Since a capacitor acts as a short circuit i₃ = 20/(5 + 4) = 9 = 20/9 A. Hence, v₃ = (20/9) × 4 = 80/9 V and v_c = 0. (ii) Under steady state conditions, capacitor acts as an open circuit and coil as a short circuit. Hence, i₂ = 20/ (5 + 7) = 20/12 = 5/3 A; v₂ = 7 × 5/3 = 35/3 V; v_L = 0. Also i₃ = 0, v₃ = 0 but v_c = 20V.

22.

Calculate the values of i2, i3, v2, v3, va, vc and vL of the network shown in Fig. 5.40 at the following times : (i) At time, t = 0 + immediately after the switch S is closed ; (ii) At time, t → ∞ i.e. in the steady state.

Answer»

(i) In this case the coil acts as an open circuit, hence i₂ = 0 ; v₂ = 0 and v_L = 20 V.

Since a capacitor acts as a short circuit i₃ = 20/(5 + 4) = 9 = 20/9 A. Hence, v₃ = (20/9) × 4 = 80/9 V and v_c = 0.

(ii) Under steady state conditions, capacitor acts as an open circuit and coil as a short circuit. Hence, i₂ = 20/ (5 + 7) = 20/12 = 5/3 A; v₂ = 7 × 5/3 = 35/3 V; v_L = 0.

Also i₃ = 0, v₃ = 0 but v_c = 20V.

Discussion

23.	Combined capacitance is equal to the A. sum of all capacitance of capacitors B. product of all the capacitance C. difference between the capacitors D. average capacitance of capacitors
Answer» A. sum of all capacitance of capacitors

Discussion

24.	If the capacitors are connected in parallel, then the potential difference across each capacitor is A. same B. different C. zero D. infinite
Answer» The Correct option is A. same

Discussion

25.	Capacitance and charge on plates are A. inversely related B. directly related C. not related at all D. always equal
Answer» B. directly related

Discussion

26.	If the plates of capacitor are oppositely charged then the total charge is equal toA. negativeB. positiveC. zeroD. infinite
Answer» The Correct option is C. zero

Discussion

27.	Area under current-time graph represents A. magnitude of charge B. dielectric C. amount of positive charge D. amount of negative charge
Answer» A. magnitude of charge

Discussion

28.	If charge stored on plates of capacitor is large, then capacitance will be A. small B. large C. zero D. infinite
Answer» The Correct option is B. large.

Discussion

29.	Hall probe is made up of A. metals B. non metals C. semiconductor D. radioactive material
Answer» C. semiconductor

Discussion

30.	For an electron, magnitude of force on it is A. BeV B. eV C. Be D. BV
Answer» The Correct option is A. BeV

Discussion

31.	When current is parallel to magnetic fields, force on conductor is A. zero B. infinite C. 2 times D. same
Answer» The Correct option is A. zero

Discussion

32.	According to the equation ‘r = (mv) ⁄(Be) ’, the faster moving particles A. move in smaller circle B. move straight C. move in bigger circle D. move randomly
Answer» C. move in bigger circle

Discussion

33.	Direction of conventional current is A. direction of neutron flow B. direction of electron flow C. direction of flow of positive charge D. same as that of electric current
Answer» C. direction of flow of positive charge

Discussion

34.	A parallel-plate capacitor is charged to 50 μC at 150 V. It is then connected to another capacitor of capacitance 4 times the capacitance of the first capacitor. Find the loss of energy.
Answer» C₁ = 50/150 = 1/3 μF ; C₂ = 4 × 1/3 = 4/3 μF Before Joining E₁ = 1/2C₁V₁² = 1/2 x (1/3) x 10^-6 x 150² = 37.5 x 10^-4J; E₂ = 0 Total energy = 37.5 × 10^{− 4} J After Joining When the two capacitors are connected in parallel, the charge of 50 μ C gets redistributed and the two capacitors come to a common potential V. V = total charge/total capacitance = 50μC/([(1/3) + (4/3)]μF) = 30V E₁ = 1/2 x (1/3) x 10^-6 x 30² = 1.5 x 10^-4J E₂= 18/2 x (4/3) x 10^-6 x 30² = 6.0 x 10^-4J Total energy = 7.5 x 10^-4J; Loss of energy = (37.5 − 7.5) × 10^{− 4} = 3 × 10^{− 2} J The energy is wasted away as heat in the conductor connecting the two capacitors.

34.

A parallel-plate capacitor is charged to 50 μC at 150 V. It is then connected to another capacitor of capacitance 4 times the capacitance of the first capacitor. Find the loss of energy.

Answer»

C₁ = 50/150 = 1/3 μF ; C₂ = 4 × 1/3 = 4/3 μF

Before Joining

E₁ = 1/2C₁V₁² = 1/2 x (1/3) x 10^-6 x 150²

= 37.5 x 10^-4J;

E₂ = 0

Total energy = 37.5 × 10^{− 4} J

After Joining

When the two capacitors are connected in parallel, the charge of 50 μ C gets redistributed and the two capacitors come to a common potential V.

V = total charge/total capacitance

= 50μC/([(1/3) + (4/3)]μF) = 30V

E₁ = 1/2 x (1/3) x 10^-6 x 30² = 1.5 x 10^-4J

E₂= 18/2 x (4/3) x 10^-6 x 30² = 6.0 x 10^-4J

Total energy = 7.5 x 10^-4J;

Loss of energy = (37.5 − 7.5) × 10^{− 4} = 3 × 10^{− 2} J

The energy is wasted away as heat in the conductor connecting the two capacitors.

Discussion

35.	Two underground cables having conductor resistances of 0.7 Ω and 0.5 and insulation resistance of 300 M Ω respectively are joind (i) in series (ii) in parallel. Find the resultant conductor and insulation resistance.
Answer» (i) The conductor resistance will add like resistances in series. However, the leakage resistances will decrease and would be given by the reciprocal relation. Total conductor resistance = 0.7 + 0.5 = 1.2 Ω If R is the combined leakage resistance, then 1/R = 1/300 + 1/600 ∴ R = 200MΩ (ii) In this case, conductor resistance is = 0.7 × 0.5/(0.7 + 0.5) = 0.3. Ω (approx) Insulation resistance = 300 + 600 = 900 M Ω

35.

Two underground cables having conductor resistances of 0.7 Ω and 0.5 and insulation resistance of 300 M Ω respectively are joind (i) in series (ii) in parallel. Find the resultant conductor and insulation resistance.

Answer»

(i) The conductor resistance will add like resistances in series. However, the leakage resistances will decrease and would be given by the reciprocal relation.

Total conductor resistance = 0.7 + 0.5 = 1.2 Ω

If R is the combined leakage resistance, then

1/R = 1/300 + 1/600

∴ R = 200MΩ

(ii) In this case, conductor resistance is

= 0.7 × 0.5/(0.7 + 0.5) = 0.3. Ω (approx)

Insulation resistance = 300 + 600 = 900 M Ω

Discussion

36.	A capacitor-type stored-energy welder is to deliver the same heat to a single weld as a conventional welder that draws 20 kVA at 0.8 pf for 0.0625 second/weld. If C = 2000 μF, find the voltage to which it is charged.
Answer» The energy supplied per weld in a conventional welder is W = VA × cos φ× time = 20,000 × 0.8 × 0.0625 = 1000 J Now, energy stored in a capacitor is (1/2) CV² ∴ W = 1/2CV² or V = √(2W/C) = √((2 x 1000)/(2000 x 10^-6)) = 1000V

36.

A capacitor-type stored-energy welder is to deliver the same heat to a single weld as a conventional welder that draws 20 kVA at 0.8 pf for 0.0625 second/weld. If C = 2000 μF, find the voltage to which it is charged.

Answer»

The energy supplied per weld in a conventional welder is

W = VA × cos φ× time = 20,000 × 0.8 × 0.0625 = 1000 J

Now, energy stored in a capacitor is (1/2) CV²

∴ W = 1/2CV² or V = √(2W/C) = √((2 x 1000)/(2000 x 10^-6)) = 1000V

Discussion

37.	The insulation resistance of a kilometre of the cable having a conductor diameter of 1.5 cm and an insulation thickness of 1.5 cm is 500 M Ω. What would be the insulation resistance if the thickness of the insulation were increased to 2.5 cm ?
Answer» The insulation resistance of a cable is First Case R = (2.3ρ/2πl)log₁₀(r₂/r₁) r₁ = 1.5/2 = 0.75 cm ; r₂ = 0.75 + 1.5 = 2.25 cm ∴ r₂/r₁ = 2.25/0.75 = 3 ; log₁₀ (3) = 0.4771 ∴ 500 = (2.3ρ/2πl) x 0.4771 ..... (i) Second Case r₁ = 0.75 cm − as before r₂ = 0.75 + 2.5 = 3.25 cm r₂/r₁ = 3.25/0.75 = 4.333 ; log₁₀ (4.333) = 0.6368 ∴ R = (2.4ρ/2πl) x 0.6368 ......(ii) Dividing Eq. (ii) by Eq. (i), we get R/500 = 0.6368/0.4771; R = 500 x 0.6368/0.4771 = 667.4MΩ

37.

The insulation resistance of a kilometre of the cable having a conductor diameter of 1.5 cm and an insulation thickness of 1.5 cm is 500 M Ω. What would be the insulation resistance if the thickness of the insulation were increased to 2.5 cm ?

Answer»

The insulation resistance of a cable is

First Case R = (2.3ρ/2πl)log₁₀(r₂/r₁)

r₁ = 1.5/2 = 0.75 cm ; r₂ = 0.75 + 1.5 = 2.25 cm

∴ r₂/r₁ = 2.25/0.75 = 3 ; log₁₀ (3) = 0.4771

∴ 500 = (2.3ρ/2πl) x 0.4771 ..... (i)

Second Case

r₁ = 0.75 cm − as before r₂ = 0.75 + 2.5 = 3.25 cm

r₂/r₁ = 3.25/0.75 = 4.333 ; log₁₀ (4.333) = 0.6368

∴ R = (2.4ρ/2πl) x 0.6368 ......(ii)

Dividing Eq. (ii) by Eq. (i), we get

R/500 = 0.6368/0.4771; R = 500 x 0.6368/0.4771 = 667.4MΩ

Discussion

38.	Find the Ceq of the circuit shown in Fig. . All capacitances are in μ F
Answer» Capacitance between C and D = 4 + 1 \|\| 2 = 14/3 μ F. Capacitance between A and B i.e. C_eq= 3 + 2 \|\| 14/3 = 4.4 μ F

38.

Find the Ceq of the circuit shown in Fig. . All capacitances are in μ F

Answer»

Capacitance between C and D = 4 + 1 || 2 = 14/3 μ F.

Capacitance between A and B

i.e. C_eq= 3 + 2 || 14/3 = 4.4 μ F

Discussion

39.	A cable is 300 km long and has a conductor of 0.5 cm in diameter with an insulation covering of 0.4 cm thickness. Calculate the capacitance of the cable if relative permittivity of insulation is 4.5.
Answer» Capacitance of a cable is C = (0.024ε_r)/log₁₀(b/a)μF/km Here, a = 0.5/2 = 0.25 cm ; b = 0.25 + 0.4 = 0.65 cm ; b/a = 0.65/0.25 = 2.6 ; log₁₀ ^2.6 = 0.415 ∴ C = (0.024 x 4.5)/0.415 = 0.26 Total capacitance for 300 km is = 300 × 0.26 = 78μF

39.

A cable is 300 km long and has a conductor of 0.5 cm in diameter with an insulation covering of 0.4 cm thickness. Calculate the capacitance of the cable if relative permittivity of insulation is 4.5.

Answer»

Capacitance of a cable is C = (0.024ε_r)/log₁₀(b/a)μF/km

Here, a = 0.5/2 = 0.25 cm ; b = 0.25 + 0.4 = 0.65 cm ; b/a = 0.65/0.25 = 2.6 ; log₁₀ ^2.6 = 0.415

∴ C = (0.024 x 4.5)/0.415 = 0.26

Total capacitance for 300 km is = 300 × 0.26 = 78μF

Discussion

40.	A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the:A. capacitance B. surface charge density on each plate C. stored energy D. electric field between the two places E. charge on each plate
Answer» C. stored energy

Discussion

41.	If dielectric slab of thickness 5 mm and εr = 6 is inserted between the plates of an air capacitor with plate separation of 8 mm, its capacitance is (a) decreased (b) almost doubled (c) almost halved (d)unaffected
Answer» Correct option (b) almost doubled

41.

If dielectric slab of thickness 5 mm and εr = 6 is inserted between the plates of an air capacitor with plate separation of 8 mm, its capacitance is (a) decreased (b) almost doubled (c) almost halved (d)unaffected

Answer»

Correct option (b) almost doubled

Discussion

42.	The capacitance of a capacitor is NOT influenced by (a) plate thickness (b) plate area (c) plate separation (d) nature of the dielectric
Answer» Correct option (a) plate thickness

42.

The capacitance of a capacitor is NOT influenced by (a) plate thickness (b) plate area (c) plate separation (d) nature of the dielectric

Answer»

Correct option (a) plate thickness

Discussion

43.	A capacitor consists of two (a) insulation separated by a dielectric(b) conductors separated by an insulator(c) ceramic plates and one mica disc (d) silver-coated insulators
Answer» Correct option (b) conductors separated by an insulator

43.

A capacitor consists of two (a) insulation separated by a dielectric(b) conductors separated by an insulator(c) ceramic plates and one mica disc (d) silver-coated insulators

Answer»

Correct option (b) conductors separated by an insulator

Discussion

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