24
May
2 batteries connected in series thru a resistor. What happens to current if ONE of the batteries is switched?
Posted by sandman49
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?
May 24th, 2010 at 10:11 pm
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
May 24th, 2010 at 10:11 pm
the flipped battery will be reversed in polarity, thus applying a backflow current!
although i dunt really get ur question?