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Three resistors, of 40 Ω, 60 Ω, and 120 Ω are connected in parallel and this parallel group
is connected with 15 Ω in series with 25 Ω. The whole system is then connected to a 120 V source.
Determine (a) the I in 25 Ω, (b) the potential drop across the parallel group,
(c) the potential across the 25 Ω, (d) the I in the 60 Ω, and (e) the I in 40 Ω.
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Resistance of the parallel group is = = = = 20 Ω.
So, now we can consider the circuit as a series of 20 Ω, 15 Ω and 25 Ω with the combined resistance of (20 Ω + 15 Ω + 25 Ω) = 60 Ω.
Hence, the electric current " I " through 25 Ω (a) is = 2 amperes.
Thus question (a) is answered.
Potential drop across the parallel group is 20*2 = 40 V (b); across the 25 Ω is 25*2 = 50 V (c).
Thus questions (b) and (d) are answered.
Solved, explained and completed.
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