Answer:
The drop voltage is 0.3 V
Explanation:
Electromotive Force EMF
When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:
[tex]\displaystyle i=\frac{\varepsilon }{R_e+R_i}[/tex]
The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:
[tex]\displaystyle i=\frac{24 }{158+2}[/tex]
[tex]\displaystyle i=\frac{24 }{160}[/tex]
i = 0.15 A
The drop voltage is the voltage of the internal resistance:
[tex]V_i = i.R_i[/tex]
[tex]V_i = 0.15*2[/tex]
[tex]\boxed{V_i = 0.3\ V}[/tex]
The drop voltage is 0.3 V
