Question

Three cells of emf 2v 1.8v and 1.5v and internal resistance 0.05 ohm, 0.7 ohm, and 1.0 ohm, respectively are joined in series. This battery is connected to an external resistor of 4.0 ohm, through a low resistnece ammeter. What would the ammeter read?

Answer 1

Answer:

I = 0.92 A

Explanation:

Three cells of emf 2v 1.8v and 1.5v and internal resistance 0.05 ohm, 0.7 ohm, and 1.0 ohm, respectively are joined in series. This battery is connected to an external resistor of 4.0 ohm, through a low resistnece ammeter

Let say Ammeter reads = I  ampere

E - Ir = IR

r = internal resistance

R = External Resistance

2 - I(0.05) + 1.8 - I(0.7)  + 1.5 (1)  = I * 4

=> 5.3 - I(1.75) = 4I

=> 5.3 = 5.75I

=> I = 5.3/5.75

=> I = 0.92 A