Chpt 3 Capacitance And Dielectrics Flashcards
3.1 Cells are connected in parallel in order to A Increase the charge available B Reduce cost of wiring C Increase the current available D Reduce the time required to fully charge them in use
Answer C Increase the current available Explanation Connecting cells in parallel increases the overall current capacity of the system The voltage remains the same but the total current that can be drawn is the sum of the individual cell currents
3.2 The combined resistance of two resistors connected in parallel is equal to A One half the resistance of one resistor B Twice the resistance of one resistor C Four times the resistance of one resistor D One fourth the resistance of one resistor
Answer A One half the resistance of one resistor (if the resistors are equal) Explanation The formula for resistors in parallel is 1/Rtotal = 1/R₁ + 1/R₂ If R₁ = R₂ then Rtotal = R₁/2 If the resistors are not equal the total resistance will be less than the smallest individual resistance
3.3 Materials which can store electrical energy are called A Magnetic materials B Semiconductors C Dielectric materials D Paper conductors
Answer C Dielectric materials Explanation Dielectric materials are insulators that can store electrical energy when placed in an electric field They are crucial components of capacitors
3.4 The dielectric constant of air is practically A More than unity B Unity C Less than unity D Zero
Answer B Unity (approximately) Explanation The dielectric constant of air is very close to 1 It’s slightly greater than 1 but often approximated as 1 in many calculations
3.5 A 1µF capacitor is charged using a constant current of 10µA for 20s What is the energy finally stored by the capacitor A 2 x 10⁻⁷ J B 2 x 10⁻⁵ J C 4 x 10⁻⁷ J D 4 x 10⁻⁵ J
Answer A 2 x 10⁻⁷ J Explanation 1 Charge (Q) = I * t = (10 x 10⁻⁶ A) * (20 s) = 2 x 10⁻⁴ C 2 Energy (E) = 1/2 * C * V² = 1/2 * Q * V We need to find V V = Q/C = (2 x 10⁻⁴ C) / (1 x 10⁻⁶ F) = 200 V 3 Energy (E) = 1/2 * (2 x 10⁻⁴ C) * (200 V) = 2 x 10⁻² J There’s a discrepancy here The options seem to have an error The correct energy would be 0.02J which is not an option
3.6 A 1000µF capacitor initially uncharged is charged by a steady current of 50µA How long will it take for the potential difference across the capacitors to reach 2.5V A 20s B 50s C 100s D 400s
Answer B 50s Explanation 1 Charge (Q) = C * V = (1000 x 10⁻⁶ F) * (2.5 V) = 2.5 x 10⁻³ C 2 Time (t) = Q / I = (2.5 x 10⁻³ C) / (50 x 10⁻⁶ A) = 50 s
3.7 What is the capacitance C of a capacitor charged at a potential difference of 60V whose quantity of charge is 3x10⁻³C A 5µF B 5µF C 1.8x10⁻⁵ F D 1.8x10⁻⁵ F
Answer None of the options are correct Explanation Capacitance (C) = Q / V = (3 x 10⁻³ C) / (60 V) = 5 x 10⁻⁵ F = 50µF
3.8 Three capacitors of capacitances 2µF 5µF and 6µF are connected in series What is the equivalent capacitance A 4/13 µF B 3/4 µF C 3µF D 13/µF
Answer A 4/13 µF (approximately) Explanation For capacitors in series 1/Ctotal = 1/C₁ + 1/C₂ + 1/C₃ 1/Ctotal = 1/2 + 1/5 + 1/6 = 23/30 Therefore Ctotal ≈ 1.3 µF The closest option is A
3.9 The capacitor stores energy and this energy is equal to the work done in A charging the capacitor B charging the battery C storing the charge D lifting a coulomb of charge
Answer A charging the capacitor Explanation The energy stored in a capacitor is directly related to the work done in charging it
3.10 What is the charge on the 1.5µF capacitor (Diagram needed for accurate calculation) A 0.87x10⁻⁵ C B 2.16x10⁻⁵ C C 3.96x10⁻⁵ C D 1.80 x 10⁻⁵ C
Answer Requires the diagram to determine the equivalent capacitance and voltage across the 1.5µF capacitor before calculating the charge (Q=CV)
3.11 The dimensions of the plates of a parallel plate capacitor are 8cm and they are separated by a distance of 2mm Calculate the capacitance if (i) air is between the plates (ii) glass of dielectric constant 5 fills the space between the plates
Answer Requires calculation using the formula C = ε₀ * A * k / d where ε₀ is the permittivity of free space A is the area k is the dielectric constant and d is the separation distance
3.12 To achieve very large capacitance what variables must be considered in designing the capacitor
Answer Increase the area of the plates (A) use a material with a high dielectric constant (k) and decrease the separation distance between the plates (d)
3.13 A charge on this capacitance of 2µF is charged to a difference of potential of 100V Find the charge having a capacitance
Answer Q = C*V = (2 x 10⁻⁶ F) * (100 V) = 2 x 10⁻⁴ C = 200µC Explanation Use the basic capacitance formula Q = CV
3.14 Define capacitance and obtain an expression for a capacitor of a parallel plate capacitor
Answer Capacitance is the ability of a component to store electrical charge For a parallel plate capacitor C = ε₀ * A * k / d
3.15 The potential difference between the plates of the capacitor is 1000V and each plate carries 250µC Calculate the capacitance and energy stored in the capacitor
Answer Capacitance (C) = Q/V = (250 x 10⁻⁶ C) / (1000 V) = 2.5 x 10⁻⁷ F = 0.25 µF Energy (E) = 1/2 * C * V² = 1/2 * (2.5 x 10⁻⁷ F) * (1000 V)² = 0.125 J
3.16 Deduce expressions for the combined capacitance of two capacitors (a) connected in series (b) connected in parallel
Answer (a) Series 1/Ctotal = 1/C₁ + 1/C₂ (b) Parallel Ctotal = C₁ + C₂
3.17 Two capacitors of 6µF are connected in series and then connected to an external source of voltage 1000V Find (i) the total capacitance (ii) the potential drop across each capacitor (iii) the charge on each capacitor
Answer (i) Ctotal = C₁C₂/(C₁+C₂) = (6µF * 6µF)/(6µF + 6µF) = 3µF (ii) Potential drop across each is half the source voltage (500V) because they are equal and in series (iii) Charge (Q) = CV = (3 x 10⁻⁶ F) * (1000 V) = 3 x 10⁻³ C = 3000µC
3.20 Two parallel wires are suspended in a vacuum system (the two charges are of opposite sign) Wires is 48µ each wires has a charge of 72 x 10⁻⁶ C the potential difference between the two wires is 48V Calculate the capacitance of 72 x 10⁻⁶ when the two charges are of opposite sign Calculate the capacitance of the parallel wires system
Answer Requires more information or clarification on the wording of the problem particularly concerning the “48µ” and the relevance of the two charge values given The calculation of capacitance for a parallel wire system involves a different formula than a parallel plate capacitor
- A parallel-plate capacitor with plates of area A and separation d is charged to a potential difference V. What is the energy density of the electric field between the plates? a) (1/2)ε₀E² b) ε₀E² c) (1/2)CV² d) QV
Answer: (a) (1/2)ε₀E² Explanation: The energy density (energy per unit volume) associated with an electric field is given by (1/2)ε₀E² where ε₀ is the permittivity of free space and E is the electric field strength.
- A dielectric material is inserted between the plates of a capacitor that is connected to a battery. What happens to the charge on the capacitor? a) It decreases. b) It increases. c) It remains the same. d) It becomes zero.
Answer: (b) It increases. Explanation: When a dielectric is inserted into a capacitor connected to a battery the capacitance increases. Since the voltage across the capacitor remains constant (due to the battery) the charge on the capacitor must increase to satisfy the equation Q = CV.
- A capacitor is charged by a battery and then disconnected. If the distance between the plates of the capacitor is increased what happens to the potential difference across the capacitor? a) It increases. b) It decreases. c) It remains the same. d) It becomes zero.
Answer: (a) It increases. Explanation: When the battery is disconnected the charge on the capacitor plates remains constant. Increasing the distance between the plates decreases the capacitance (C = ε₀A/d). Since Q = CV the potential difference (V) must increase to maintain a constant charge.
- Two capacitors one with capacitance C₁ and the other with capacitance C₂ are connected in parallel to a battery. What is the equivalent capacitance of this combination? a) C₁C₂/(C₁ + C₂) b) C₁ + C₂ c) (C₁ + C₂)/2 d) √(C₁C₂)
Answer: (b) C₁ + C₂ Explanation: The equivalent capacitance of capacitors connected in parallel is simply the sum of their individual capacitances.
- Which of the following is NOT a factor that affects the capacitance of a capacitor? a) The area of the plates b) The distance between the plates c) The charge on the plates d) The dielectric material between the plates
Answer: (c) The charge on the plates Explanation: Capacitance is a property of the physical configuration of the capacitor and does not depend on the amount of charge stored on it. The other factors listed all influence the capacitor’s ability to store charge.
- What is the unit of dielectric constant? a) Farad (F) b) Volt (V) c) Coulomb (C) d) None it is dimensionless.
Answer: (d) None it is dimensionless. Explanation: The dielectric constant is a ratio of the permittivity of a material to the permittivity of free space. Since both permittivities have the same units the dielectric constant is dimensionless.
- What is the primary function of a capacitor in an electrical circuit? a) To resist the flow of current. b) To store electrical energy. c) To amplify electrical signals. d) To convert AC to DC.
Answer: (b) To store electrical energy. Explanation: Capacitors store energy in the electric field that is created between their charged plates.
- A capacitor with a capacitance of 10 μF is charged to a potential difference of 100 V. How much energy is stored in the capacitor? a) 0.05 J b) 0.5 J c) 5 J d) 50 J
Answer: (b) 0.5 J Explanation: The energy stored in a capacitor can be calculated using the equation U = (1/2)CV². In this case: U = (1/2)(10 × 10⁻⁶ F)(100 V)² = 0.5 J
- The dielectric strength of a material refers to: a) The ability of a material to store electrical energy. b) The amount of charge a material can hold. c) The resistance of a material to the flow of current. d) The maximum electric field a material can withstand before it breaks down.
Answer: (d) The maximum electric field a material can withstand before it breaks down. Explanation: Dielectric strength is a measure of a dielectric material’s ability to withstand a high electric field without becoming conductive.
- Why does inserting a dielectric material between the plates of a capacitor increase its capacitance? a) The dielectric material increases the area of the plates. b) The dielectric material decreases the distance between the plates. c) The dielectric material reduces the electric field strength between the plates. d) The dielectric material increases the voltage across the plates.
Answer: (c) The dielectric material reduces the electric field strength between the plates.Explanation: When a dielectric is inserted its molecules become polarized creating an electric field that opposes the external field. This effectively reduces the net electric field allowing more charge to be stored at the same potential difference.
- A dielectric material with dielectric constant K is inserted between the plates of a parallel plate capacitor. What happens to the capacitance? a) It decreases by a factor of K. b) It increases by a factor of K. c) It stays the same. d) It decreases by a factor of 1/K.
Answer: (b) It increases by a factor of K. Explanation: The presence of a dielectric material between the plates of a capacitor increases the capacitance by a factor equal to the dielectric constant (K) of the material. This is because the dielectric reduces the electric field strength between the plates allowing more charge to be stored at the same potential difference.
