Bipolar and health insurance

Yes, insurance is very expensive. Your employer must have had a pretty great plan for Cobra to be $1,000. By law, the cost for Cobra cannot exceed 102% the cost of the original plan. They aren't tacking on a bunch of extra costs. I don't think people realize how much their employers spend on health insurance.

I know that the lower cost health plan has a lot less coverage, but you need to decide if you are actually going to use $6,000 worth of health service that the lower cost plan doesn't offer that the higher one does. That's the extra cost per year. You could use that if something isn't covered. If under this lower-cost plan your deductible is over $1,200 for yourself or over $2,400 for your family, consider putting that extra money you're saving into an Health Savings Account, which is a pre-tax account that you can use for non-covered health expenses, like co-pays, prescriptions, durable medical equipment, etc.



I think the bottom line is that you need to figure out how much you spend on medical expenses and if that extra $500 in premiums every month is going to save you money or if you'd rather have that in your pocket. Look at what doctors are in the network (are your doctors there?), what prescriptions are covered (are yours on the list and how much would you have to pay for them if they aren't? Could you go generic?), and what percentage is your responsibility after the deductible is met? Also, check what your maximum out-of-pocket expenses are on each plan, ie. when the insurance company starts to pay everything.

Just don't wait too long to decide or you'll have problems either getting Cobra or with per-existing conditions with a new insurance company.

I hope that helps. Best of luck.

What I have found so far is that there seems to be nothing affordable or practical for us. We cannot afford 500bux/mo for insurance, nor the medications or doc visits. I am pretty much screwed bc I live in the U.S.A. and there is nothing to help with medical expenses or insurance premiums.