A transistor in a common Emitter configuration is used as a switch, and it is in the ON position. The output voltage ${V_o}$ is:

Hint: You should have some basic idea about the transistor, types of transistor and how a transistor works. What is collector, base and the emitter of a transistor and about different types of configuration such as common base configuration, common emitter configuration and common collector configuration.

Complete step by step solution:
 When the transistor is in the saturation area and the switch is in the ON position, the IC is extremely high and the ${V_ \circ }$ is nearly nil or zero.
When a transistor is used as a switch in a CB (common base) configuration, the input emitter current needed to activate the switch (to drive the transistor to saturation) is nominally equal to the output collector being switched.
The input voltage to operate the switch in CC (common collector) is the same as the supply voltage provided by the main supply or the battery we connected to the circuit.
However, in CE (common emitter), the input switching signal (current, $i.e.$ base current or voltage) used to drive the switch is insignificant as compared to the output current or voltage that is switched. As a result, CE (common emitter) is more convenient to use as a switch.

Note:The transistor in an ideal switch should only be in one of two states: off or on. When there is no bias voltage or when the bias voltage is less than \[0.7{\text{ }}V\], the transistor is turned off. When the base is saturated, the switch is turned on to allow collector current to flow freely.