A battery X of emf 6volt and internal resistance 2(ohms) is connected in series with a battery Y of emf 4volt and internal resistance 8(ohms) so that the two emf act to oppose each other, calculate the terminal P.D of X and Y
I = (E1-E2)/(r1+ r2) = (6-4)/(2+8) = 0.2A.
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.
but i dont know how to solve it
To calculate the terminal P.D (Potential Difference) of batteries X and Y when connected in series, we need to apply Kirchhoff's Voltage Law (KVL). According to KVL, the sum of the potential differences in a closed loop is equal to zero.
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.
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