Doubtly
@Doubtly · Joined Feb 2021
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- Distinguish between a potentiometer and a voltmeter. PotentiometerVoltmeter 1. A potentiometer is used to determine the emf of a cell, potential difference and internal resistance.1. A voltmete... 1
- On what factors does the internal resistance of a cell depend? The internal resistance of a cell depends on : Nature of the electrolyte : The greater the conductivity of the electrolyte, the lower is the... 1
- A voltmeter has a resistance 30 Ω. What will be its reading, when it is connected across a cell of emf Data: E = 2V, r = 10 Ω, R = 30 Ω The voltmeter reading, V = IR = ( E/R+r ) R = ( 2/30+10 ) 30 = ( 2/40 ) 30 = 1.5 V 1
- When two cells of emfs. E1 and E2 are connected in series so as to assist each other, their balancing length Data : l₁ = 2.7m (cells assisting), l₂ = 0.3 m (cells opposing) E₁ + E₂ = Kl₁ and E₁ – E₂ = Kl₂ ∴ fracE₁+E₂E₁-E₂=fracK l₁K l₂ ∴ fracE₁E₂=fra... 1
- The emf of a cell is balanced by a length of 120 cm of potentiometer wire. When the cell is shunted by a resistance Data :R = 10 Ω, l₁ = 120 cm, l₂ = 120 – 20 = 100 cm r = R( fracl₁-l₂l₂ ) = 10 ( 120-100/100 ) = 2 Ω The internal resistance of the cell is 2... 1
- A potential drop per unit length along a wire is 5 × 10-3 V/m. If the emf of a cell balances against Data: K = 5 × 10⁻³ V/m , L = 216 cm = 216 × 10⁻² m E = KL ∴ E = 5 × 10⁻³ × 216 × 10⁻² = 1080 × 10⁻⁵ = 0.01080V The emf of the cell is 0.0108... 1
- The resistance of a potentiometer wire is 8 Ω and its length is 8 m. A resistance box and a 2 V Data: R = 8 Ω, L = 8 m, E = 2 V, K = 1 µV/mm = 1 × frac10⁻⁶ ~V10⁻³ ~m = 10⁻³ V/m K = V/L=fracE R(R+R_B) L^(prime) where R_B is the resistanc... 1
- Find the equivalent resistance between the terminals F and B in the network shown in the figure below given Applying Kirchhoff’s voltage law to loop FGHF, we get, – 10I1 – 10(I1 – I2) + 10(I – I1) + 10(I – I1) = 0 ∴ – 10I1 – 10I1 + 10I2 + 10I – 10I... 1
- A voltmeter has a resistance of 100 Ω. What will be its reading when it is connected across a cell of emf 2 V Data: E = 2V, r = 20 Ω, R = 100 Ω The voltmeter reading, V = IR V = ( 2/100+20 )100 = 200/120=10/6 = 1.667 V. 1
- A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge Answer: (D) Constant, increases, decreases Charge: the capacitor is isolated after charging, so the charge has nowhere to go. Q stays consta... 1
- Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio. (A) 1 : 1 Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio 1 : 1 1
- A parallel plate capacitor has circular plates of radius 8 cm and plate separation 1mm. (A) 1.78 × 10⁻⁸ C 2
- If the difference between the radii of the two spheres of a spherical capacitor is increased The capacitance of a spherical capacitor is C = 4πε₀ (a b/b-a) where a and b are the radii of the concentric inner and outer conducting shel... 1
- A metal plate is introduced between the plates of a charged parallel plate capacitor. Suppose the parallel-plate capacitor has capacitance C₀, plates of area A and separation d. Assume the metal sheet introduced has the same a... 1
- The safest way to protect yourself from lightening is to be inside a car. Justify. There is danger of lightning strikes during a thunderstorm. Because trees are taller than people and therefore closer to the clouds above, t... 1
- A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. Assume a parallel-plate capacitor, of plate area A and plate separation d is filled with a dielectric of relative permittivity (dielectric c... 1
- Two charges of magnitudes -4Q and +2Q are located at points (2a, 0) and (5a, 0) respectively. The sphere of radius 4a encloses only the negative charge Q₁ = -4Q. The positive charge Q₂ = +2Q being located at a distance of 5a from the... 1
- A 6 µF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected Data: C = 6 µF = 6 × 10⁻⁶ F = C₁, V = 300 V C₂ = 3 µF The electrostatic energy in the capacitor = 1/2 Cv² = 1/2 (6 × 10⁻⁶)(300)² = 3 × 10⁻⁶... 1