Question

A battery of e.m.f. 12 V and internal resistance 0.5 Ω is connected to a 9.5 Ω resistor through a key. The ratio of potential difference between the two terminals of the battery, when the key is open to that when the key is closed, is ______.

पर्याय

• 1.05

• 1

• 0.95

• 1.1

Answer 1

A battery of e.m.f. 12 V and internal resistance 0.5 Ω is connected to a 9.5 Ω resistor through a key. The ratio of potential difference between the two terminals of the battery, when the key is open to that when the key is closed, is 1.05.

Explanation:

Given: EMF of the battery (E) = 12 V

Internal resistance (r) = 0.5 Ω

External resistance (R) = 9.5 Ω

When the key is open, no current flows in the circuit.

V_open = E

= 12 V

When the key is closed, current flows in the circuit. The total resistance in the circuit is:

R_total = R + r

= 9.5 + 0.5

= 10 Ω

The current flowing through the circuit is given by Ohm’s law:

I = `E/R_"total"`

= `12/10`

= 1.2 A

The potential difference across the battery terminals when current is flowing is:

`V_"closed"` = E − Ir

= 12 − (1.2 × 0.5)

= 12 − 0.6

= 11.4 V

∴ `V_"open"/V_"closed" = 12/11.4` 

= 1.05