This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the questions you've provided: 15. The equivalent resistance of two identical resistors in parallel is 10 Ω. The equivalent resistance of the same resistors in series would be Step 1: Let the resistance of each identical resistor be R. For two resistors in parallel, the equivalent resistance R_p is given by: (1)/(R_p) = (1)/(R) + (1)/(R) (1)/(R_p) = (2)/(R) R_p = (R)/(2) Step 2: We are given that R_p = 10 \, . So, we can find R: 10 \, = (R)/(2) R = 2 × 10 \, = 20 \, Step 3: For the same two resistors in series, the equivalent resistance R_s is given by: R_s = R + R R_s = 2R Step 4: Substitute the value of R: R_s = 2 × 20 \, = 40 \, The correct option is C. 40 16. The capacitance of a parallel plate is 20µF in air and 42µF when a dielectric is placed between its plate. The dielectric constant of the material is Step 1: The capacitance of a parallel plate capacitor in air (or vacuum) is C_0 = 20 \, . Step 2: The capacitance with a dielectric material is C = 42 \, . Step 3: The dielectric constant (or _r) is defined as the ratio of the capacitance with the dielectric to the capacitance in air: = (C)/(C_0) Step 4: Substitute the given values: = 42 \, 20 \, = 2.1 The correct option is C. 2.1 17. How much work is done in moving a 2.0 C of charge against a potential difference of 12 Volts? Step 1: The work done W in moving a charge Q against a potential difference V is given by the formula: W = Q × V Step 2: Substitute the given values: Q = 2.0 \, C and V = 12 \, V. W = 2.0 \, C × 12 \, V W = 24 \, J The correct option is C. 24 J 18. A current through a 50 Ω resistor is 2 A. How much power is dissipated as heat? Step 1: The power P dissipated as heat in a resistor is given by Joule's law: P = I^2 R where I is the current and R is the resistance. Step 2: Substitute the given values: I = 2 \, A and R = 50 \, . P = (2 \, A)^2 × 50 \, P = 4 \, A^2 × 50 \, P = 200 \, W The correct option is C. 200 W 19. Kirchhoff's first law is the law of conservation of ________. Kirchhoff's first law, also known as Kirchhoff's current law (KCL), states that the algebraic sum of currents entering a node (or a junction) is equal to the algebraic sum of currents leaving it. This law is a direct consequence of the conservation of charges. Charges cannot accumulate at a node; they must flow in and out. The correct option is D. charges 20. The amount of work done in bringing a unit positive charge from infinity to a point in an electric field is ________. The amount of work done in bringing a unit positive charge from infinity to a point in an electric field is the definition of electric potential at that point. The correct option is A. Electric potential 1. For a parallel plate capacitor, if the charge stored on a capacitor plate is large, the capacitance will be ________. Step 1: The capacitance C of a capacitor is defined as the ratio of the charge Q stored on its plates to the potential difference V across them: C = (Q)/(V) Step 2: For a given capacitor, its capacitance C is a constant value determined by its physical dimensions (plate area, distance between plates) and the dielectric material. It does not depend on the charge stored or the potential difference across it. If the charge stored (Q) increases, the potential difference (V) across the plates also increases proportionally, keeping the ratio C constant. Therefore, the capacitance itself does not change based on the charge stored. However, the question implies a relationship where a "large charge stored" might lead to a "large capacitance". This is a common misconception. Capacitance is an intrinsic property. If we consider the design of a capacitor, to store a large charge for a given voltage, it must have a large capacitance. If the question implies a design choice, then a capacitor designed to store a large charge (at a typical operating voltage) would be a large capacitance capacitor. Given the multiple-choice options, the most likely intended answer is that a capacitor designed to store a large charge has a large capacitance. The correct option is A. large 2. A capacitor connected to a 24 V battery has a charge of 4 x 10⁻³ C. Its capacitance is ________. Step 1: The capacitance C of a capacitor is given by the formula: C = (Q)/(V) where Q is the charge stored and V is the potential difference. Step 2: Substitute the given values: Q = 4 × 10^-3 \, C and V = 24 \, V. C = 4 × 10^-3 \, C24 \, V C = (1)/(6) × 10^-3 \, F C ≈ 0.1666... × 10^-3 \, F C ≈ 0.0001666... \, F Step 3: Convert the capacitance to microfarads (), where 1 \, = 10^-6 \, F. C = 0.1666... × 10^-3 \, F = 166.6... × 10^-6 \, F C ≈ 167 \, The correct option is C. 167 \, 3. A pico farad is ________ of a farad. Step 1: The prefix "pico" (p) represents 10^-12. Step 2: Therefore, a pico farad (pF) is 10^-12 of a farad. The correct option is B. 10^-12 4. The resistance of a 120 V light bulb that draws a current of 1.25 A is ________. Step 1: According to Ohm's Law, the resistance R is given by: R = (V)/(I) where V is the voltage and I is the current. Step 2: Substitute the given values: V = 120 \, V and I = 1.25 \, A. R = 120 \, V1.25 \, A R = 96 \, The correct option is B. 96 \, 5. A certain wire has a resistance R. Another wire, exactly the same except having twice the diameter, has the resistance ________. Step 1: The resistance R of a wire is given by the formula: R = (L)/(A) where is the resistivity of the material, L is the length, and A is the cross-sectional area. Step 2: The cross-sectional area A of a circular wire is given by A = r^2, where r is the radius. Since diameter d = 2r, then r = d/2, so A = ((d)/(2))^2 = ( d^2)/(4). Step 3: Substitute the area into the resistance formula: R = (L)/( d^2)4 = (4 L)/( d^2) Step 4: For the second wire, the diameter is d' = 2d. All other parameters (, L) are the same. Let its resistance be R'. R' = (4 L)/( (d')^2) = (4 L)/( (2d)^2) = (4 L)/( (4d^2)) = (1)/(4) ( (4 L)/( d^2) ) Step 5: From Step 3, we know that R = (4 L)/( d^2). So, R' = (1)/(4) R The correct option is D. (1)/(4)R 6. The current in the 5 Ω resistor is ________. Step 1: The resistors 5 \, and 10 \, are connected in series. The total equivalent resistance R_eq for series resistors is the sum of individual resistances: R_eq = R_1 + R_2 R_eq = 5 \, + 10 \, = 15 \, Step 2: A potential difference V = 60 \, V is applied across the series combination. According to Ohm's Law, the total current I flowing through the circuit is: I = (V)/(R_eq) I = 60 \, V15 \, I = 4 \, A Step 3: In a series circuit, the current is the same through all components. Therefore, the current in the 5 \, resistor is 4 \, A. The correct option is A. 4 A 7. The potential difference across the 5 Ω resistor is ________. Step 1: From the previous question (Question 6), we found that the current I flowing through the series circuit is 4 \, A. Step 2: To find the potential difference V_1 across the 5 \, resistor (R_1), we use Ohm's Law: V_1 = I × R_1 Step 3: Substitute the values: I = 4 \, A and R_1 = 5 \, . V_1 = 4 \, A × 5 \, V_1 = 20 \, V The correct option is A. 20 V 8. A current of 9 A flows when a 120 V battery is connected across a 12 Ω resistor. The battery has an internal resistance of ________. Step 1: For a circuit with a battery having an electromotive force (EMF) E and internal resistance r, connected to an external resistor R, the current I flowing in the circuit is given by: I = ER + r Step 2: We are given: I = 9 \, A, E = 120 \, V, and R = 12 \, . We need to find r. Rearrange the formula to solve for r: I(R + r) = E R + r = EI r = EI - R Step 3: Substitute the values: r = 120 \, V9 \, A - 12 \, r = 13.333... \, - 12 \, r = 1.333... \, Rounding to one decimal place, r ≈ 1.3 \, . The correct option is A. 1.3 \, 9. A 10 Ω resistor and a 5 Ω resistor are connected in parallel. Their equivalent resistance is ________. Step 1: For two resistors connected in parallel, the equivalent resistance R_eq is given by the formula: (1)/(R_eq) = (1)/(R_1) + (1)/(R_2) Step 2: Substitute the given values: R_1 = 10 \, and R_2 = 5 \, . (1)/(R_eq) = (1)/(10 \, ) + (1)/(5 \, ) Step 3: Find a common denominator and add the fractions: (1)/(R_eq) = (1)/(10 \, ) + (2)/(10 \, ) (1)/(R_eq) = (3)/(10 \, ) Step 4: Invert the result to find R_eq: R_eq = (10)/(3) \, R_eq ≈ 3.33 \, The correct option is B. 3.3 \, 10. Electric charge occurs only in separate parcels ________. Electric charge is quantized, meaning it exists only in discrete units. The smallest unit of charge is the elementary charge, e, which is the magnitude of the charge of an electron or a proton. This value is approximately 1.6 × 10^-19 \, C. Therefore, electric charge occurs only in separate parcels of ± 1.6 × 10^-19 \, C. The correct option is A. of ± 1.6 × 10^-19 C 11. Two charges of + 2.0 x 10⁻⁶ C that are 3.0mm apart repel each other with a force of ________. Step 1: Coulomb's Law describes the force between two point charges: F = k (|q_1 q_2|)/(r^2) where F is the electrostatic force, k is Coulomb's constant (9 × 10^9 \, Nm^2/C^2), q_1 and q_2 are the magnitudes of the charges, and r is the distance between their centers. Step 2: Identify the given values: q_1 = q_2 = 2.0 × 10^-6 \, C r = 3.0 \, mm = 3.0 × 10^-3 \, m k = 9 × 10^9 \, Nm^2/C^2 Step 3: Substitute the values into Coulomb's Law: F = (9 × 10^9 \, Nm^2/C^2) (2.0 × 10^-6 \, C)(2.0 × 10^-6 \, C)(3.0 × 10^-3 \, m)^2 F = (9 × 10^9) 4.0 × 10^-129.0 × 10^-6 \, N F = (9 × 10^9) × (4)/(9) × 10^-12 - (-6) \, N F = 4 × 10^9 - 12 + 6 \, N F = 4 × 10^3 \, N The correct option is A. 4 × 10^3 N 12. An electric force of 1.0 x 10⁻⁵ N acts on a charge of 5.0 x 10⁻¹⁰ C that is between two parallel metal plates 4.0mm apart. The potential difference between the plates is ________. Step 1: The electric field E between two parallel plates is related to the force F on a charge q by: F = qE So, the electric field strength is: E = (F)/(q) Step 2: Substitute the given values: F = 1.0 × 10^-5 \, N and q = 5.0 × 10^-10 \, C. E = 1.0 × 10^-5 \, N5.0 × 10^-10 \, C E = 0.2 × 10^5 \, N/C = 2.0 × 10^4 \, N/C Step 3: The potential difference V between two parallel plates separated by a distance d in a uniform electric field E is given by: V = E × d Step 4: The distance d = 4.0 \, mm = 4.0 × 10^-3 \, m. V = (2.0 × 10^4 \, N/C) × (4.0 × 10^-3 \, m) V = 8.0 × 10^4-3 \, V V = 8.0 × 10^1 \, V = 80 \, V The correct option is C. 80 V 13. Ten millimeters from a certain charge, its electric field is 10 kV/m. The magnitude of the field 20mm from the charge is ________. Step 1: The electric field E due to a point charge Q at a distance r is given by: E = k (Q)/(r^2) where k is Coulomb's constant. This shows that E is inversely proportional to r^2 (E (1)/(r^2)). Step 2: We are given E_1 = 10 \, kV/m at r_1 = 10 \, mm. We need to find E_2 at r_2 = 20 \, mm. We can set up a ratio: (E_2)/(E_1) = (k Q)/(r_2^2)k (Q)/(r_1^2) = (r_1^2)/(r_2^2) = ((r_1)/(r_2))^2 Step 3: Substitute the values: (E_2)/(10 \, kV/m) = (10 \, mm20 \, mm)^2 (E_2)/(10 \, kV/m) = ((1)/(2))^2 (E_2)/(10 \, kV/m) = (1)/(4) Step 4: Solve for E_2: E_2 = (1)/(4) × 10 \, kV/m E_2 = 2.5 \, kV/m The correct option is A. 2.5 kV/m 14. The potential difference across any number of capacitors in parallel equals the ________. When capacitors are connected in parallel, they are all connected across the same two points in the circuit. This means that the potential difference (voltage) across each individual capacitor is the same and equal to the total potential difference across the parallel combination. The correct option is C. potential difference across each capacitor What's next? Send 'em! 📸