Answer :
When batteries X and Y are in series, the terminal voltage for X is 5V and Y is 0V. When battery Y is reversed, the terminal voltage for X changes to 5.8V and Y to 4.8V.
Let's first calculate the total emf when the batteries are in series and acting in the same direction:
emf_total = 6 V + 4 V
= 10 V
The total internal resistance (r_total) is: r_total
= 2 Ω + 8 Ω
= 10 Ω
The circuit has a 10 Ω external resistor.
The total resistance in the circuit (R_total) is: R_total = r_total + external resistance
= 10 Ω + 10 Ω
= 20 Ω
The current (I) in the circuit can be calculated using Ohm's Law (V = IR):
I = emf_total / R_total
= 10 V / 20 Ω
= 0.5 A
Next, the terminal potential difference (V_terminal) for each battery is:
For Battery X: V_terminal_X = emf_X - I x r_X
= 6 V - 0.5 A x 2 Ω
= 5 V
For Battery Y: V_terminal_Y = emf_Y - I x r_Y
= 4 V - 0.5 A x 8 Ω
= 0 V
When Battery Y is reversed, the emf values oppose each other:
emf_total = 6 V - 4 V
= 2 V
Total circuit resistance remains the same (20 Ω), so the new current is: I_new = emf_total / R_total
= 2 V / 20 Ω
= 0.1 A
The new terminal potential differences are:
For Battery X:
V_terminal_X = emf_X - I_new x r_X
= 6 V - 0.1 A x 2 Ω
= 5.8 V
For Battery Y:
V_terminal_Y = emf_Y + I_new x r_Y
= 4 V + 0.1 A x 8 Ω
= 4.8 V