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Question
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
Options
5J
10J
20J
30J
Solution
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" = 6/6`
I = 1A
Heat dissipated through 4Ω resistor is given by
H = I2 × R × t
H = I2 × 4 × 5
H = 20J
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