Question

cell of EMF 2v and internal resistance 1.2 ohm is connected to an ammeter of resistance 0.8 ohm and two resistor of 4.5 ohm and 9 ohm find the potential difference across the 4.5 resistor ​

Answer 1

Explanation:

EMF of cell = 2V

Internal resistance r = 1.2 ohm

Resistance of ammeter = 0.8 ohm

Resistance of parallel combination Rp = 1 / (1/4.5 +1/9) = 3 ohm

Resistance in series Rs = resistance of ammeter + Rp =0.8 + 3 =3.8 ohm

Total external resistance R = 3.8 ohm

a) Now, ammeter reading I = (emf of cell)/ R + r I = 2/(3.8 + 1.2).

Ans a). I = 0.4 A

b) p.d across terminals of cell V = I * R = 0.4 * 3.8 = 1.52 V

Ans b). 1.52 V

c).

Let current from 4.5ohm be I1 and from 9ohm be I2

Then, I = I1 + I2 0.4 = I1 + I2 ; I2 = 0.4 - I1

In parallel p.d. is same across all resistances.

So, I1 * 4.5 = I2 * 9 I1 = (9 * I2)/4.5 I1 = {9 * (0.4 - I1)}/4.5 I1 = 0.8/3 A Therefore, p.d. across 4.5 ohm = I1 * 4.5 =0.8/3 * 4.5 = 1.2 V

Ans c). 1.2 V