[SOLVED] Load calculations when connected in series and parallel

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Abhishek_Anand

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I have a load of 10W, 75 Ohms connected to a DC supply. Now if I add another load of the same rating in parallel to this load, keeping the supply same. What will be the new load across the supply.
What if I add it in series instead of parallel..?
How does load rating change when connected in series and parallel?
 

When you are connecting the same 75 ohm load connected in parallel, the current from the voltage source will get equally divided in the two loads. For your case, 10W @ 75 ohm needs 0.36Amps from the voltage source. If you add one more load parallel, another 0.36Amps also will flow from the voltage source. This will effectively make a 0.73Amps current flow on the load which is equivalent to 20W power from the source.

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Load rating will not change. Instead the load will share the total power so that it will not exceed the maximum rating of the load.
 
Sorry. Did not get you.

As you said, adding another load will change the effective resistance and hence the current being drawn from the source.
Hence now I have 20W of load.

What do you mean by the last line..?

"Load rating will not change. Instead the load will share the total power so that it will not exceed the maximum rating of the load."
 

For a particular load resistor, say 75ohms 10W, it is always fixed. If you dissipate more than 10W, the resistor will burn. Only you can do is to reduce the power rating by sharing the power.

If you are connecting the same load in parallel, the current drawn will double-hence the power delivered by the source will double. If you connecting same load in series, the current drawn will get half since the resistance is doubled. so the power delivered by the source will get half.
 
When added in parallel, the net load as seen by the battery would be 20W.
When added in series, the net load as seen by the battery would be 5W.

Recall that in series the current would reduce to half, and since power delivered by the battery = VI, the power is reduced to half ie. 5W
5W is being supplied by the battery and it is being shared equally by the two resistors, therefore the resistors would consume 2.5W each.
 
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