What Led to the Fall of the Sikh Empire?

During ranjit singh the power of the sikhs was on climax.ranjit singh gave special attention to the organization of army he had built a state whose main base was army the main foundation of a military state is efficient leadership.after the death of ranjit singh the army become cause of dissolution of sikh Kingdome. After ranajit singh due to the weak successors all power went to the khalsa army.in the 5 yrs of the death of ranjit Singh the number of soldiers increased 3 times and salary was also increased and due to the decreasing income of the state this situation became unbearable. Soldiers formed their own Panchayats through them they start making important decisions like military campaigns,supporting the state's clamaint etc. thus due to the absence of skilled leadership basic power of state became the reason for decline of state. And other important reason for decline of sikh state was internal conspiracy. Thus the moral decline of army ,internal conspiracy,falling economic situation make sikh state a victim of the british policy of grab.

1. Are there any alternatives to a Netanyahu-led government in Israel and, if so, what are they?

Israel heads to the polls on March 17th, 2015, and Benjamin Netanyahu, though still the front-runner, appears to be vulnerable. I look at the challenge facing Labor, Netanyahu's main opposition, the Labor has governed Israel for only 8. Four elements are needed for Labor to win:1. The incumbent--Netanyahu--has to be vulnerable in his own right, which Netanyahu is.2. Labor needs to shift the debate toward the economy, where it has an electoral advantage. 3. Labor has to present a credible alternative on national security, it's electoral weakness, and4. Labor needs a credible leader to replace the incumbent. nFor the full post.

2. What tv to buy? LED or Plazma?

PLAZMAAAAAA!!(= WOOOOOHOOOOO!!

3. how to put led to indicate the signal is functioning

As Olin points out, it's hard to tell what you are really asking. I am going to assume that you want an indicator LED to light up whenever the optocoupler in ON. In that case, your circuit should look like the one below:The "limit switch", when closed, should tied the INPUT signal to 3. 3V and light up the indicator LED and the optocoupler's internal LED. The OUTPUT signal goes to an input of your PIC microcontroller.The value of the resistors will depend on the optocoupler and the LED you choose to use.

4. wiring multiple led flashlights to single power source [closed]

Let me start out with a summary of my understanding of the problem.The goal is to eliminate buying lots of batteries by centralizing the power source used for a system of between 15 and 30 lanterns, each separated by 30 feet, for a distance run that can be from 500 to 1000 feet long. These lanterns, in prior use, used a set of (3) AAA batteries and are very simple lanterns using one resistor of $22:Omega$ and (24) parallel-wired LEDs. (Parallel-wiring like that does not allow the LEDs to share current as well as they may, but it's how it was done and these are just lanterns so variations between the 24 LEDs is fine for this application.) Each lantern requires approximately:$$I_lantern=frac4.5:textrmV-2:textrmV22:Omegale 125:textrmmA$$Some options to consider:Well, those are some directions that could be used without needing to modify the lanterns. If you are open to completely new lantern designs, then there are still more options to consider.Since Passerby dragged my attention back here, might as well expose the computation for computing the voltage drop at the far-end of the line given periodic loads along a line. (I've no clue if there is a commonly used symbol for $fracOmega1000:textrmft$, so I am just calling it $k_wire$ for now. $D$ will be the total distance of the run. $N$ will be the number of periodic loads, which divide up the total distance into $N$ segments of wire. $I_load$ is the per-unit load. The first load is one segment downstream of the power source. The last load is at the very end.)$$beginalign* V_drop &= sum_i=1^N left[I_i=I_loadcdotleft(N-i1

ight)

ight]cdotleft[R_i=2cdotfrack_wirecdot D1000cdot N

ight] V_drop &= I_loadcdotleftcdotfrack_wirecdot D1000cdot N

ight] sum_i=1^N left(N-i1

ight) V_drop &= I_loadcdotleftcdotfrack_wirecdot D1000cdot N

ight]cdotleft[ sum_i=1^N N-sum_i=1^N isum_i=1^N1

ight] V_drop &= I_loadcdotleftcdotfrack_wirecdot D1000cdot N

ight]cdotleft[ N^2-sum_i=1^N iN

ight] V_drop &= I_loadcdotleftcdotfrack_wirecdot D1000cdot N

ight]cdotleft[ fracN^2N2

ight] V_drop &= left[I_total=I_loadcdotfracN12

ight]cdot left[R_total=2cdot k_wirecdotfracD1000

ight] endalign*$$This proves that it comes out very close to the natural presumption of using an average current that is one-half the total current (for large $N$) times the total wire resistance, out and back. But it never hurts to do the math to test common sense. (And to also see that it's not exactly the same, too.)