I include the following unusually simple MCQ just because it appeared in IIT Screening 2005 question paper. The interesting thing is that some of the students with top ranks in the class may omit such trivial questions because of sheer suspicion. Here is the question:

**Find the current through 2Ω**

**(a) 0**

**(b) 1 A**

**(c) 2 A**

**(d) 4 A**

Both poles of the 10V battery are connected to one end of the 2Ω resistor. Similarly, both poles of the 20V battery are connected to the other end of the 2Ω resistor. So, there are no continuous paths for the currents to flow through the 2Ω resistor and the current is zero.

**Now, suppose the above circuit is modified as shown in the adjoining figure in which the points A and B are grounded. Assuming that the batteries have negligible internal resistance, what will be the current through the 2Ω resistor?**

The two batteries are now connected (through the ground connection) in series with the 2Ω resistor. But, the battery voltages are in opposition so that the net voltage across the 2Ω resistor is 20V – 10 V = 10 V. Therefore, the current is 10V/2Ω = 5A.

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