2 batteries connected in series thru a resistor. What happens to current if ONE of the batteries is switched?

Question

Two batteries connected in series through a resistor are bucking eachother. (Current and battery voltages given). If one of the batteries is switched (flipped), what is the new current?

Answer 1

bucking each other >> connected in opposing coupling

.< I1... +| V1|- ... -|V2|+...> I2 ... |

....................... ............. ........ |

........ <<< R >>>> .................. |

current in circuit = I2- I1 = (V1 - V2)/R = I (given)

so R = (V1 - V2) / I ............. (1)

==================

when battery V2 is flipped

new voltage (V1+V2)

new current (i) = [V1+V2] /R

put the value of R from (1)

i = [V1+V2] * I /[V1 - V2] >>>>>> answer ----- (2)

now battery are in tune >> supportively coupled

note >if you flip V1 then only direction of (i) will change not its magnitude

===============

if V1 = 15 V, V2 = 5 V, I = 2 A

then on flipping V2, you get

i = (15+5)*2/[15 - 5] = 4 A

typically current doubled for just chosen values

Answer 2

the flipped battery will be reversed in polarity, thus applying a backflow current!

although i dunt really get ur question?