Determine the terminal voltage of each battery in both cases

Determine the terminal voltage of each battery in both cases R = 10.0 omega, R_a=0.8 omega,R_b=2.6 omaga (R_a and R_b are the internal resistancess of the respective batteries.)

Solution

a)

Applying Kirchoff\'s loop rule

E₁-IR-E₂-IR_a-IR_b=0

10-10I-5-0.8I-2.6I=0

13.4I =5

I =0.373 A

So Voltage drop across 10 V battery is

V₁ = E₁-IR_a =10-0.373*0.8

V₁ =9.7 Volts

So Voltage drop across 5 V battery is

V₂ = E₂+IR_b =5+0.373*2.6

V₂=5.97 Volts

b)

Applying Kirchoff\'s loop rule

E₁-IR-E₂-IR_a-IR_b=0

5-10I-10-0.8I-2.6I=0

13.4I =-5

I =-0.373 A

So Voltage drop across 5 V battery is

V₁ = E₁-IR_a =10-(-0.373)*0.8

V₁ =10.3 Volts

So Voltage drop across 10 V battery is

V₂ = E₂+IR_b =5-0.373*2.6

V₂=4.03 Volts