A battery with an internal resistance of 2 Ω and an emf of 4.0 V is connected in series to a load resistance and the terminal voltage falls to 3.8 V. What current is flowing in the circuit, and what is the value of the load resistance?

Concept:

Battery: 

• ​The battery is used for the conversion of energy from one form to another.
• In these devices, one terminal becomes positively charged while the other becomes negatively charged.
• Therefore, an electromotive force is work done on a unit of electric charge.

Emf: ​Electromotive force 

• It is defined as the electric potential produced by either an electrochemical cell or by changing the magnetic field

• The formula, E = ir + iR, where E =  emf, i = current flowing in the circuit, r = internal resistance, R = external resistance connected in series.

Calculation:

Given,

The emf of battery, E = 4.0 V

The internal resistance of the battery, r = 2.0 Ω 

The resistance connected to the circuit, R = ?

The terminal voltage, V_T = 3.8 V 

The terminal voltage is given as, V_T = E - ir

3.8 = 4 - 2i

2i =4 - 3.8

2i = 0.2

\(i=\frac{0.2}{2}=0.1A\)

Apply Kirchoff's voltage law in the above circuit, E = ir + iR

\(i=\frac{E}{R+r}\)

\(R+r=\frac{E}{i} \)

\(R+2=\frac{4}{0.1}=40\)

R = 40 - 2 = 38 Ω