13 Potentiometer & It's Applications | Current Electricity Class 12 | JEE Mains & Advanced
Mohit Tyagi
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13 Potentiometer & It's Applications | Current Electricity Class 12 | JEE Mains & Advanced
123 445 просмотров · 6 лет назад
Mohit Tyagi
2,22 млн подписчиков
123 445 просмотров · 6 лет назад
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Topics covered in this Current Electricity Class 12 lecture by ABJ sir IITD with timestamp:
00:00 - 14:05 Potentiometer: The potentiometer is an instrument used to measure the unknown voltage by comparing it with the known voltage. It can be used to determine the emf and internal resistance of the given cell and to compare the emf of different cells. ABJ Sir also draw its structure. The basic principle of the potentiometer is that the potential drop across any section of the wire will be directly proportional to the length of the wire, provided the wire is of a uniform cross-sectional area and a uniform current flows through the wire.
14:05 - 18:21 Current Electricity Problem 1: Based on the potentiometer. In this example, there is a potentiometer with a driver cell of 20V and internal resistance of 5 ohms. The length of the Wire is 20 m with a resistance of 100 ohms. ABJ Sir, calculate the current in this circuit, potential difference across AB and maximum voltage, the potential gradient
18:37 - 21:02 Working principle of potentiometer with comparison to a weighing machine.
21:08 - 26:58 How to use Potentiometer to find the potential difference: A potentiometer is a device used to measure the potential difference between two points. The potentiometer works on the principle that when a constant current flows through a wire of uniform cross-sectional area, the potential difference(E) between its two points is directly proportional to the length(l) of the wire between the two points, so E is directly proportional to l.
26:58 - 31:35 Important point: If there is no null point on the potentiometer wire: ABJ Sir explains this point with the help of a diagram and tells us that if there is no null point on the wire, then the unknown potential difference is more than the potential difference of wire AB.
31:55 - 40:45 Current Electricity Problem 2: Based on the potentiometer - To find unknown EMF of a cell. In this example, there is a potentiometer with a driver cell and internal resistance. These two primary cells are connected to a galvanometer via a two way key. The galvanometer is connected to a high resistance box and then to the Jockey. The Jockey has to find the point of null deflection in the galvanometer (N1and N2). Using the ratio of both EMF relation with length of the null point, we can find the value of unknown EMF.
41:03 - 47:28 Current Electricity Problem 3: Based on the potentiometer - To find Current in the Circuit. In this example, there is a potentiometer with a driver cell and internal resistance. These two primary cells are connected to a galvanometer via a two way key (Different arrangement from the last circuit arrangement). The galvanometer is connected to a high resistance box and then to the Jockey. The Jockey has to find the point of null deflection in the galvanometer (N1and N2). Using the ratio of both EMF relation with length of the null point, we can find the value of unknown EMF. Then to find the value of current is also easy.
48:31 - 01:01:01 Current Electricity Problem 4: Based on the potentiometer - To find internal resistance of a battery. In this example, there is a potentiometer with a driver cell and internal resistance. The galvanometer is connected to a high resistance box and then to the Jockey. The Jockey has to find the point of null deflection in the galvanometer . Using the ratio of both EMF relation with length of the null point, we can find the value of unknown EMF, current in the secondary circuit. So using this method we can find out the value of internal resistance of the battery.