The potentiometer wire PQ is 100 cm long and its resistance is 2r, where r is the internal resistance of the battery. PC is equal to :
Correct answer: B. 75 cm
- A. 25 cm
- B. 75 cm
- C. 50 cm
- D. 40 cm
Explanation
The potentiometer works by comparing the EMF of a cell with a potential drop across a known length of wire. Since the wire PQ is 100 cm long, and its resistance is 2r, where r is the internal resistance of the battery, the potential drop is proportional to the length of the wire. The balance point PC is found by equating the potential difference across PC with the EMF of the unknown source. Thus, the balance point is at 75 cm, where the potential difference equals the EMF, taking into account the internal resistance and the total resistance of the wire.
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Steady current relates charge flow to potential difference, resistance and resistivity, including how temperature changes a conductor's resistance. Sources with internal resistance have terminal voltage below their emf when delivering current, and maximum power output occurs under a specific load condition rather than at zero resistance.
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