EAMCET (Engineering Entrance) 2010 Questions on Capacitors

“Non-violence is the greatest force at the disposal of mankind. It is mightier than the mightiest weapon of destruction devised by the ingenuity of man”

– Mahatma Gandhi

Today we will discuss three questions which appeared in EAMCET (Engineering Entrance) 2010 question paper. Questions (1) and (2) are from electrostatics while question no.3 is from transient phenomena.

(1) Two capacitors of capacities 1 μF and C μF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacity C in microjoule is

(1) 1800

(2) 1600

(3) 14400

(4) 7200

The effective capacitance of the series combination is 1×C/(1+C) μF = C/(1+C) μF.

[Remember that when capacitors C₁, C₂, C₃,….etc are connected in series, the effective capacitance C_eff is given by the reciprocal relation, 1/C_eff = 1/C₁ + 1/C₂ + 1/C₃ + …. etc.]

We retain the unit of capacitance as μF for convenience.

Since the charge carried by the combination is 80 μC, we have (from Q = CV with usual notations)

80 = [C/(1+C)]×120

[We have substituted the charge in microcoulomb itself since the capacitance is in microfarad].

This gives 80 C + 80 = 120 C from which C = 2 μF

When capacitors are in series, all capactors carry the same charge. Therefore, the charge on the 2 μF capacitor is 80 microcoulomb.

The energy U of the 2 μF capacitor is given by

U = Q²/2C = 80²/(2×2) = 1600 microjoule.

[Since Q is in microfarad and C is in microcoulomb, the energy is in microjoule].

(2) The potential difference between two parallel plates is 10⁴ volts. If the plates are separated by 0.5 cm, the force on an electron between the plates is

(1) 32×10^–13 N

(2) 0.32×10^–13 N

(3) 0.032×10^–13 N

(4) 3.2×10^–13 N

This is a very simple question. The electric field (E) between the plates is given by

E = V/d where V is the potential difference and d is the separation between the plates

The force F on the electron (of caharge e = 1.6×10^–19 coulomb) is given by

F = Ee = Ve/d = (10⁴×1.6×10^–19) /(0.5×10^–2) = 3.2×10^–13 N

(3) A capacitor of capacity 0.1 μF connected to a resistor of 10 MΩ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given log₁₀ 2 = 0.3010)

(1) 2 sec

(2) 0.693 sec

(3) 0.5 sec

(4) 1 sec

If V₀ is the initial voltage across the capacitor of capacitance C, the voltage V at time t during the discharge through the resistance R is given by

V = V₀ e^–t/RC where e is the base of natural logarithms.

Therefore V/V₀ = e^–t/RC

Since V/V₀ = ½ in the question, we have

½ = e^–t/RC

Or, e^t/RC = 2

Taking logarithm, t/RC = ln 2

Therefore, t = RC ln 2 = 10×10⁶×0.1×10^–6×ln 2

If you remember that ln 2 = 0.693, you can write the answer immediately as 0.693 sec.

[You should definitely know that the relation between natural logarithm (with base e) and common logarithm (base 10) of quantity x is

ln_e x = 2.303 log₁₀ x

So ln 2 = 2.303×0.3010 = 0.693]