Two cells of emfs approximately 5v and 10v
WebTwo cells of emf's approximately 5 V and 1 0 V are to be accurately compared using a potentiometer of length 4 0 0 cm. A. The battery that runs the potentiometer should have voltage of 8 V. B. The battery of potentiometer can have a voltage of 1 5 V and R adjusted … WebQuestion: Two cells of emfs approximately 5V and 10V are to be accurately compared using a potentiometer of length 400 cm. (a) the battery that runs the potentiometer should have voltage of 8V (b) the battery of potentiometer can have a voltage of 15V and R adjusted so that the potential drop across the wire slightly exceeds 10V (c) the first portion of 50 cm …
Two cells of emfs approximately 5v and 10v
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WebAug 16, 2024 · The potential drop along the wires of potentiometer should be greater than emfs of cells. In a potentiometer experiment, the emf of a cell can be measured if the potential drop along the potentiometer wire is more than the emf of the cell to be determined. Here, values of emfs of two cells are given as 5 V and 10 V, therefore, the … WebDec 10, 2024 · The electromotive force (EMF) is the maximum potential difference between two electrodes of a galvanic or voltaic cell. This quantity is related to the tendency for an element, a compound or an ion to acquire (i.e. gain) or release (lose) electrons. For example, the maximum potential between Zn and Cu of a well known cell.
WebTwo cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm. 5. A metal rod of length 10 cm and a rectangular cross … WebJul 17, 2024 · Answer/Explanation. 2. Two batteries of ε 1 and ε 2 (ε 2 > ε 1) and internal resistance r 1 and r 2 respectively are connected in parallel as shown in figure. [NCERT Exemplar] (a ) The equivalent emf ε eq of the two cells is between ε 1 and ε 2 i.e. ε 1 < ε eq < ε 2. (b) The equivalent emf ε eq is smaller than ε 1.
WebAug 30, 2024 · Two cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer of length 400 cm. WebJul 18, 2024 · Two cells of emf 10V and 5V having internal resistance 2ohm and 1ohm respectively are connected in parallel as shown.The potential through 10ohm resistor wou...
WebThe potential drop along the wires of potentiometer should be greater than emfs of cells. In a potentiometer experiment, the emf of a cell can be measured if the potential drop along …
WebTwo cells of emfs approximately 5V and 10V are to be accurately compared using a poteniometer ... used for comparing resistances and not voltages. date filled bear claw pastryWebTwo cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm. A. The battery that runs the potentiometer should have a … bivalvia circulatory systemWebOct 26, 2024 · Two cells of emf’s E1 and E2 (E1 > E2 ) are connected in a potentiometer circuit so as to assist each other. asked Mar 1, ... Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm. asked Jul 30, 2024 in Physics by Anukriti bharti (38.3k points) bivalving cast meansWebTwo cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm.a) The battery that runs the potentiometer sh... bivalvia factsWebOct 5, 2024 · In case the two cells are identical each of emf E=5v and internal resistance r=2ohm calculate the voltage across the external resistance R=10ohm See answers Advertisement Advertisement Mousmi25 Mousmi25 Answer: V=4.54 volt. Explanation: Eeq = (E1r2 + E2r1)/r1×r2 = ( 5×2 + 5×2 ) / 2×2 date filed meaningWebMay 16, 2024 · Two cells of emf’s approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm. a. The potentiometer is usually used for comparing resistances and not voltages. b. The first portion of 50 cm of the wire itself should have a potential drop of 10V. c. bivalving a castWebThis will be equal to the EMF of each cell. This means that the total EMF provided by the combination is also 5 Volts. (ii) Since there are 4 cells with internal resistance 0.5Ω connected in parallel. The equivalent internal resistance is given by the formula. 1 r e q = 1 r 1 + 1 r 2 + 1 r 3 + 1 r 4. bivalving cast