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------------------------------------------------ A battery X with an e.m.f. of 6 V and an internal resistance of 2 ohms is in series with a battery Y with an e.m.f. of 4 V and an internal resistance of 8 ohms, with both e.m.f.s acting in the same direction. A 10-ohm resistor is connected to the batteries.

1. Calculate the terminal p.d. of each battery.
2. If battery Y is reversed so that the e.m.f.s now oppose each other, what is the new terminal p.d. of X and Y?

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