Question

If three identical cell each of EMF 1.5V and internal resistance 0.5Ω are connected in series the what would the net EMF's and net internal resistance of the combination.​

Answer 1

Correct option is

A

7.5A

E

eff

=1.5×5

= 7.5 V

r

eff

=0.2×5

=1Ω

for maximum current flow load resistance =0Ω

⇒E

eff

=Ir

eff

⇒I=

1

7.5

=7.5A

Answer 2

Given :

The emf of the cell = 1.5V

Internal resistance (r) = 0.5Ω

From graph we observe that

V = I

=> resistance of the conductor (R) = V / I

(∵ from ohm's law)

=> 1Ω

So,

the total resistance of the current path =R+r

=> 1 + 0.5Ω

=> 1.5Ω

From ohm's law :

the current in the circuit (i) = emf / total resistance

=> 1.5/1.5 A

=> 1A

the voltage across the terminals = emf of cell − voltage drop across internal resistance

=> emf − (i × r)

=> 1.5−(1 × 0.5)

=> 1.0V