QUES: The casualties were horrible, especially for the attacking Prussian forces. A grand total of 20,163 German troops were killed, wounded or missing in action during the August 18 battle. The French losses were 7,855 killed and wounded along with 4,420 prisoners of war (half of them were wounded) for a total of 12,275. While most of the Prussians fell under the French Chassepot rifles, most French fell under the Prussian Krupp shells. In a breakdown of the casualties, Frossard's II Corps of the Army of the Rhine suffered 621 casualties while inflicting 4,300 casualties on the Prussian First Army under Steinmetz before the Pointe du Jour. The Prussian Guards Infantry Divisions losses were even more staggering with 8,000 casualties out of 18,000 men. The Special Guards Jäger lost 19 officers, a surgeon and 431 men out of a total of 700. The 2nd Guards Infantry Brigade lost 39 officers and 1,076 men. The 3rd Guards Infantry Brigade lost 36 officers and 1,060 men. On the French side, the units holding St. Privat lost more than half their number in the village.

What was the grand total of German casualties and those missing in action?
What is the answer?
ANS: 20,163
QUES: An antenna (plural antennae or antennas), or aerial, is an electrical device which converts electric power into radio waves, and vice versa. It is usually used with a radio transmitter or radio receiver. In transmission, a radio transmitter supplies an electric current oscillating at radio frequency (i.e. a high frequency alternating current (AC)) to the antenna's terminals, and the antenna radiates the energy from the current as electromagnetic waves (radio waves). In reception, an antenna intercepts some of the power of an electromagnetic wave in order to produce a tiny voltage at its terminals, that is applied to a receiver to be amplified.

What process associated with antennas produces a high frequency alternating current?
What is the answer?
ANS: transmission
QUES: For air dielectric capacitors the breakdown field strength is of the order 2 to 5 MV/m; for mica the breakdown is 100 to 300 MV/m; for oil, 15 to 25 MV/m; it can be much less when other materials are used for the dielectric. The dielectric is used in very thin layers and so absolute breakdown voltage of capacitors is limited. Typical ratings for capacitors used for general electronics applications range from a few volts to 1 kV. As the voltage increases, the dielectric must be thicker, making high-voltage capacitors larger per capacitance than those rated for lower voltages. The breakdown voltage is critically affected by factors such as the geometry of the capacitor conductive parts; sharp edges or points increase the electric field strength at that point and can lead to a local breakdown. Once this starts to happen, the breakdown quickly tracks through the dielectric until it reaches the opposite plate, leaving carbon behind and causing a short (or relatively low resistance) circuit. The results can be explosive as the short in the capacitor draws current from the surrounding circuitry and dissipates the energy.

Of what order is the breakdown field strength for mica dielectric capacitors?
What is the answer?
ANS:
100 to 300 MV/m