Question

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.​

Answer 1

i. 6.923 V

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

Step-by-step 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;

`(E)/(R_(balance)) = (V)/(R_(cell))`

Where:

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

`R_(balance)` = Resistance of 75 cm of potentiometer wire = 0.75×10 = 7.5 Ω

`R_(cell)` = Resistance of the cell in the circuit = 3 Ω

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

`(E)/(7.5) = (9)/(3)`

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V