Find the heat developed per minute in each of the resistors.

1.	Find the heat developed per minute in each of the resistors.
Answer» Correct Answer - `540 J, 480J, 960J` Equivalent resistance of the circuit is `(8 xx 4)/(8 + 4) xx 1 = (8)/(3) xx 1 = (11)/(3) Omega` current drawn from the battery is `1 = (11V)/((11//3) Omega) = 3A` Since the current through `1 Omega` tresistor is `3A` so heat produced in this resistor in `1` minut `(= 60` second ) is `H = l^(2)Rt = (3)^(2) xx 1 xx 60 = 540 J` current through `8 Omega` resistor `l_(1) = (3)/(8 + 4) xx 4 = 1A` Head produced in `8 Omega` resistor in `1` minut `H_(1) = l_(1)^(2) R_(1)t = (1)^(2) xx 8 xx 60 = 480 J` Current through `4 Omega` resistor `l_(2) = 1 - l_(1) = 3 - 1 = 2A` Current through `4 Omega` resistor in `1` minute `H_(2) = l_(2)^(2) R_(2) t = (2)^(2) xx 4xx 60 = 960 J`

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Find the heat developed per minute in each of the resistors.

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