total resistance=10 ohm. Current= (6-4)/10= .4 amp.
voltage drops on internal=.4*2=.8 for X
= .4*4=1.6 for Y
Now polarity. Because current is reversed in X, terminal voltage=4+.8
terminal voltage on Y= 6-1.6 Volts
voltage drops on internal=.4*2=.8 for X
= .4*4=1.6 for Y
Now polarity. Because current is reversed in X, terminal voltage=4+.8
terminal voltage on Y= 6-1.6 Volts
E1-I*r1 = 6 - 0.2 * 2 = 5.6V. = Terminal voltage 0f E1(X).
E2+I*r2 = 4 + 0.2 * 8 = 5.6v. = Terminal voltage of E2(Y).
The terminal voltage of E2(Y) is greater than 4V., because current flows INTO it.
In this case, the closed loop consists of battery X, battery Y, and the internal resistances of both batteries. Let's denote the terminal P.D of battery X as Vx and the terminal P.D of battery Y as Vy.
Applying KVL, we get:
Vx - Vy - (internal resistance of X) * (current) - (internal resistance of Y) * (current) = 0
Now, we need to express the current in terms of the terminal P.D's and the internal resistances. The current can be calculated using Ohm's Law (V = IR).
For battery X:
Vx = (emf of X) - (internal resistance of X) * (current)
Vx = 6V - 2Ω * (current)
For battery Y:
Vy = (emf of Y) - (internal resistance of Y) * (current)
Vy = 4V - 8Ω * (current)
Next, we equate the expressions for Vx and Vy:
6V - 2Ω * (current) - 4V + 8Ω * (current) = 0
Now, we can solve the equation and calculate the value of the current. Let's simplify the equation:
6V - 4V = 2Ω * (current) + 8Ω * (current)
2V = 10Ω * (current)
Dividing both sides by 10Ω:
(current) = (2V) / (10Ω)
(current) = 0.2A
Now, we can substitute the value of the current back into the expression for Vx or Vy to calculate the terminal P.D.
For battery X:
Vx = 6V - 2Ω * (0.2A)
Vx = 6V - 0.4V
Vx = 5.6V
Therefore, the terminal P.D. of battery X is 5.6 volts.
For battery Y:
Vy = 4V - 8Ω * (0.2A)
Vy = 4V - 1.6V
Vy = 2.4V
Therefore, the terminal P.D. of battery Y is 2.4 volts.