Question

Two coils of resistances `3 Omega and 6 Omega` are connected in series across a battery of `emf 12 V`. Find the electrical energy consumed in `1` minute in each resistance when these are connected in series.

Answer 1

Total resistance of the circuit, `R = 3 Omega + 6 Omega = 9 Omega`
Current in the circuit, `I = (V)/(R ) = (12 V)/(9 Omega) = (4//3) A`
Since the resistances are in series, same current flows in each resistance.
Electric energy consumed by `R_1 (=3 Omega)` in 1 minute, i.e.,
`W_1 = I^2 R_1 t = (4//3 A)^2 (3 Omega)(60 s) = 320 J` `(1 min = 60 s)`
Electric energy consumed by in `R_2 (= 6 Omega) = (4//3 A)^2 (6 Omega)(60 s) = 640 J`.