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Question
Two resistors of 4 Ω and 6 Ω are connected in parallel. The combination is
connected across a 6 volt battery of negligible resistance. Calculate:
( i) The power supplied by the battery,
(ii) The power dissipated in each resistor.
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Solution
Total resistance of resitors of 4 Ω and 6 Ω in parallel is:
RP = `[1/4 + 1/6]^-1 = 2.4 Omega`
Given voltage supply= 6 volt
(i) Power supplied by the battery, P = `"V"^2/"R" = (6 xx 6)/2.4` = 15 watt
(ii) Power dissipated in 4 Ω resistor, P1 = `"V"^2/"R" = (6 xx 6)/4` = 9 watt
Power dissipated in 6 Ω resistor, P1 `"V"^2/"R" = (6 xx 6)/6` = 6 watt
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