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In an electrical circuit, two resistors of `2 Omega and 4 Omega` respectively are connected in series to a `6 V` battery. The heat dissipated by the `4 Omega` resistor in `5 s` will be :A. 5 JB. 10 JC. 20 JD. 30 J |
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Answer» Correct Answer - C Given, resistors, `R_(1) = 2 Omega " and " R_(2) = 4 Omega` Voltage, `V = 6 V_(1)` Time, t = 5s Resistance, `R = 4 Omega` Heat dissipated, H = ? Resistor, `R_(s) = R_(1) + R_(2) = 2 + 4 = 6 Omega` Current, `I = (V)/(R) = (6)/(6) = 1A`. Heat dissipated `H = I^(2)Rt = 1 xx 4 xx 5 = 20 J` |
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