# Battery and charger question

Discussion in 'Homework Help' started by techx, Sep 21, 2009.

1. ### techx Thread Starter New Member

Sep 21, 2009
14
0
I have a question about a battery charger setup. R in the diagram is a variable resistor. I'm trying to determine what the value of R is under 3 circumstances. 1st, a charging current flows at +4amps. 2nd, a power of 25w is delivered to the battery(.035ohm and 10.5v) I forgot to label the right side of the diagram battery to the right of the nodes. and lastly, what is the value of R with a voltage of 11v present at the battery terminals. Can this be explained better than the picture.

for 1) i used KVL

0 = -13 + 10.5 - (0.02+0.035) - R*i1 ===> i1=4 and 0.02 and 0.035 can be combined cuz of series so I got

R = 0.68 ohms

for 2) I used R = (V^2)/P = (13^2)/25 = 6.76 ohm - 0.02 ohm for the resistor on charger side == 6.74 ohm

3) us not solved yet, but assuming I can use the same situation? Unless I'm suppose to take overall voltage meaning 13-10.5 for (2) and 13-11 for (3) ? Can someone shed some light?

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2. ### Mu86neer Member

May 1, 2009
23
0
Solution:
KVL==> Vs = Vb+I(R1+R+R3)
R = 1/I(Vs + I (R1+R3)-Vb).....equ.1
There are two conditions by which R has to be adjusted to adapt the situation:
(i) recall equ. 1 at Vb = 10.5v
R = 1/4(13-10.5-4(0.02+0.035) = 0.57 ohm
R=0.57 ohm
(ii) recall equ. 1 at Vb = 10.5v
R = 1/4(13-11-4(0.02+0.035) = 0.57 ohm
R=0.445 ohm
GOOD LUCK