Question

In the circuit represented by the diagram the 24.0 V battery has a terminal voltage of 21.2 V. What is the internal resistance of the battery?
A. 2.80 Ω
B. 1.43Ω
C. 0.700 Ω
D. 0.221

Answer 1

Answer:

Topic: Current Electricity

Class XII - CBSE

We are given a circuit having a battery with internal resistance 'r', connected to a load resistance 'R'. Also we are given that Terminal Voltage is 21.2 V.

The basic definition of Terminal Voltage is:

• The output voltage of any electrical device which is measured at their terminal ends.

Formula used for calculating Terminal Voltage is:

→ Terminal Voltage ( V ) = EMF of the cell - I(r)

Here, EMF is the voltage provided by the battery and 'r' is the internal resistance offered by the battery.

According to our question,

→ Terminal Voltage = 21.2 V

→ EMF of the cell = 24 V

→ I = 4 A

→ r = ?

Substituting in the formula we get,

⇒ 21.2 = 24 - 4(r)

⇒ 4(r) = 24 - 21.2

⇒ 4r = 2.8

⇒ r = 2.8 / 4 = 0.7 Ω

Hence the internal resistance offered by the battery in the given circuit is 0.7 Ω.

Answer 2

Answer:

$\bold\green{Internal\:resistance=C)0.700 \Omega }$

__________________________

Solution given below:-

Given that,

Terminal voltage = 21.2 V

Voltage offered by battery = 24.0 V

Current (I) = 4.0 A

Internal resistance (r) =?

Formula used:-

Terminal Voltage=Voltage (Battery) - Ir

$\mapsto\tt 21.2 = 24 - 4r$

$\mapsto\tt 24 - 4r = 21.2$

$\mapsto\tt - 4r = 21.2 - 24$

$\mapsto\tt - 4r = -2.8$

$\mapsto\tt r = \frac{-2.8}{-4}$

$\mapsto\tt r = 0.7 \Omega$

$\therefore Internal\:resistance =C)0.7000 \Omega$