Q1: If four 10 k resistors are in series with a single 20 k resistor and one of the parallel resistors shorts, the voltage across the other parallel resistors ________.

ANS:B - decreases

Here again, I almost forgot: As related to my last response.

I*R = V. So let's just use a random current value for the proof of say 6mA.

= (1/10*4) = 2.5K (1/10*3) = 3.33K.

= 6mA*2.5Kohm = 15 volts.

6mA*3.3Kohm = 19.8 volts, so I believe the answer may be that current decreases when one of the resistors burns out.

If the same voltage were applied to the circuit then current would definitely decrease. You can look at the question either way but the result is the same.

V = I*R.