Asked by ronald

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
Answered by bobpursley
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
Answered by ronald
the answer is 5.6volts
Answered by ronald
but i dont know how to solve it
Answered by henry2,
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.

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