A battery of internal resistance 4Ohm is connected to network of resistanc as shown .in order tgat tge maximum power can be delivered to the network ,tge value of R in ohm should be
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It is simple.
The given circuit is a balanced Wheatstone bridge. So the wire containing 6 ohm will be eliminated.
R,R,R are in series while R,R and 4R are in series.
So, R' = R + R + R = 3R
R'' = R + R + 4R = 6R
R' and R'' are in parallel
Effective Resistance = 1/3R + 1/6R = 3/6R = 6R/3 = 2R
For delivering maximum power, internal resistance = external resistance
Given, internal resistance = 4 ohm
So, 2R = 4
R = 2 ohm --> your required answer
Hope This Helps You!
The given circuit is a balanced Wheatstone bridge. So the wire containing 6 ohm will be eliminated.
R,R,R are in series while R,R and 4R are in series.
So, R' = R + R + R = 3R
R'' = R + R + 4R = 6R
R' and R'' are in parallel
Effective Resistance = 1/3R + 1/6R = 3/6R = 6R/3 = 2R
For delivering maximum power, internal resistance = external resistance
Given, internal resistance = 4 ohm
So, 2R = 4
R = 2 ohm --> your required answer
Hope This Helps You!
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