In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be

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20J Explanation - Since 2Ω and 4Ω are connected in series so their net resistance is "R"_"net" = "R"_1 + "R"_2 "R"_"net" = 2 + 4 Rnet = 6Ω Current (I) in the circuit is given by "I" = "V"/("R"_1 + "R"_2) "I" = $\frac{6}{6}$ I = 1A Heat dissipated through 4Ω resistor is given by H = I2 $\times$ R $\times$ t H = I2 $\times$ 4 $\times$ 5 H = 20J

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