Please help me with this question ASAP. In a a potentiometer circuit, a 1m long potentiometer wire PQ of resistance 10 ohms is connected in series with a cells of e. m.f. 9V with internal resistance 3 ohms Calculate:

i The p. d. across PQ
ii. The e. m.f. of a cell which has a balance point of 75cm.

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

Where:

E = e.m.f. of the balance point cell

= Resistance of 75 cm of potentiometer wire  = 0.75×10 = 7.5 Ω

= Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

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