Kirchhoff’s second rule requires
$\text{emf}-\text{Ir}-{\text{IR}}_{1}-{\text{IR}}_{2}=0$ . Rearranged, this is
$\text{emf}=\text{Ir}+{\text{IR}}_{1}+{\text{IR}}_{2}$ , which means the emf equals the sum of the
$\text{IR}$ (voltage) drops in the loop.
Applying kirchhoff’s rules
By applying Kirchhoff’s rules, we generate equations that allow us to find the unknowns in circuits. The unknowns may be currents, emfs, or resistances. Each time a rule is applied, an equation is produced. If there are as many independent equations as unknowns, then the problem can be solved. There are two decisions you must make when applying Kirchhoff’s rules. These decisions determine the signs of various quantities in the equations you obtain from applying the rules.
When applying Kirchhoff’s first rule, the junction rule, you must label the current in each branch and decide in what direction it is going. For example, in
[link] ,
[link] , and
[link] , currents are labeled
${I}_{1}$ ,
${I}_{2}$ ,
${I}_{3}$ , and
$I$ , and arrows indicate their directions. There is no risk here, for if you choose the wrong direction, the current will be of the correct magnitude but negative.
When applying Kirchhoff’s second rule, the loop rule, you must identify a closed loop and decide in which direction to go around it, clockwise or counterclockwise. For example, in
[link] the loop was traversed in the same direction as the current (clockwise). Again, there is no risk; going around the circuit in the opposite direction reverses the sign of every term in the equation, which is like multiplying both sides of the equation by
$\mathrm{\u20131.}$
[link] and the following points will help you get the plus or minus signs right when applying the loop rule. Note that the resistors and emfs are traversed by going from a to b. In many circuits, it will be necessary to construct more than one loop. In traversing each loop, one needs to be consistent for the sign of the change in potential. (See
[link] .)
When a resistor is traversed in the same direction as the current, the change in potential is
$-\text{IR}$ . (See
[link] .)
When a resistor is traversed in the direction opposite to the current, the change in potential is
$+\text{IR}$ . (See
[link] .)
When an emf is traversed from
$\u2013$ to + (the same direction it moves positive charge), the change in potential is +emf. (See
[link] .)
When an emf is traversed from + to
$\u2013$ (opposite to the direction it moves positive charge), the change in potential is
$-$ emf. (See
[link] .)
Unlike transverse waves such as electromagnetic waves, longitudinal waves such as sound waves cannot be polarized. ... Since sound waves vibrate along their direction of propagation, they cannot be polarized
Astronomy
A proton moves at 7.50×107m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength?
An American traveler in New Zealand carries a transformer to convert New Zealand’s standard 240 V to 120 V so that she can use some small appliances on her trip.
The potential difference between any two points on the surface is zero that implies È.Ŕ=0, Where R is the distance between two different points &E= Electric field intensity. From which we have cos þ =0, where þ is the angle between the directions of field and distance line, as E andR are zero. Thus
MAHADEV
sorry..E and R are non zero...
MAHADEV
By how much leeway (both percentage and mass) would you have in the selection of the mass of the object in the previous problem if you did not wish the new period to be greater than 2.01 s or less than 1.99 s?