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To locate null point deflection battery ...

To locate null point deflection battery key `(K_1)` ios pressed before the galvanometer `(K_2)` Explanation why?

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To locate the null point deflection in a potentiometer experiment, it is essential to understand the sequence of pressing the keys (K1 and K2) in the circuit. Here is the step-by-step explanation: ### Step 1: Understanding the Circuit Configuration In the experiment, we have a battery connected to a potentiometer wire, a key K1 (which connects the battery), a galvanometer with key K2, and a jockey used to find the null point. ### Step 2: Pressing Key K1 First, we press key K1. This connects the battery to the potentiometer wire, allowing a stable current to flow through the wire. It is crucial to establish a steady current in the potentiometer wire before making any measurements. ...
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To locate null point, deflection battery key K_(1) is pressed before the galvanometer key K_(2) . Explain why?

Assertion To locate null defiection the battery key (K_(1)) is pressed first and then the galvanom eter key (K_(2)) Reason if first K_(2) is pressed and then as soon as K_(1) is pressed current suddenly try to increase so due to self induction a large stopping emf is generated in galvanometer which may damage the glavanometer .

The galvanometer deflection , when key K_(1) is closed but K_(2) is open equals theta_(0) (see figure ) On closing k_(2) also and adjusting R_(2) to 5Omega , the deflection in galvanometer becomes (theta_(0))/(5) the resistnce of the galvanometer is then , given by [Neglect the internal resistance of battery]:

The resistances P,Q,R and S in the bridge shown are adjusted such that the deflection in the galvanometer G is zero when both the keys K_(1) and K_(2) are inserted. The galvanometer will show a momentary deflection if

Two coils C_(1) and C_(2) are kept coaxially as shown the coil C_(1) is connected to a battery and the coil C_(2) is connected to a galvanometer. The deflection in galvanometer can be increased by

In the arrangement shown in figure when the switch S_2 is open, the galvanometer shown no deflection for l=L//2 . When the switch S_2 is closed, the galvanometer shown no deflection for l=5L//12 . The internal resistance (r) of 6V cell, and the emf E of the other battery are respectively

In the given circuit, calculate the ratio of currents through the battery if (1) Key K_(1) is pressed, K_(2) open and then (2) Key K_(2) is pressed, key K_(1) is opened

As shown in the figure initially switch k1 is open and k2 is closed and deflection of galvanometer is x. Now both switch are closed then the deflection in galvanometer is x/5 (given R_H= 5Omega) . The resistance of galvanometer is

A moving coil galvanometer has resistance 50 Omega and it indicates full deflection at 4 m A current. A voltmeter is made using this galvanometer and a 5k Omega resistance. The maximum voltmeter, will be close be :

Two students ‘X’ and ‘Y’ perform an experiment on potentiometer separately using the circuit given: Keeping other parameters unchanged, how will the position of the null point be affected it (i) X’ increases the value of resistance R in the set-up by keeping the key K_1 closed and the key K_2 open? (ii) Y’ decreases the value of resistance S in the set-up, while the key K_2 remain open and the key K_1 closed? Justify.

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