can someone help showing the steps please​

Answer:

  a) 20 Ω

  b) 1 A

  c) 10 V on each

  d) 1/3 A through each

  e) 10 W in R1, 3 1/3 W in the others

  f) 20 W

Explanation:

The three 30 Ω resistors are in parallel with each other, so will have an equivalent resistance of (30 Ω)/3 = 10 Ω.

(a) The total circuit resistance is 10 Ω + 10 Ω = 20 Ω.

__

(b) All of the current supplied by the battery passes through R1. The current supplied by the battery is ...

  I = V/R = (20 v)/(20 Ω) = 1 A

__

(c) The voltage drop on R1 is (1 A)(10 Ω) = 10 v. Subtracting that from the battery voltage gives the voltage drop on the other resistors:

  20 v -10 v = 10 v . . . . drop on R2, R3, R4

__

(d) The current through resistors R2, R3, R4 is ...

  I = V/R = (10 v)/(30 Ω) = 1/3 A

__

(e) Power can be computed several ways. Here, it may be easiest to use ...

  P = VI

For R1, that is P1 = (10 v)(1 A) = 10 W

For R2-R4, that is P2-P4 = (10 v)(1/3 A) = 10/3 W = 3 1/3 W

__

(f) The total power dissipated is the total power supplied:

  P = VI = (20 v)(1 A) = 20 W

Adding up the power in the resistors, we get ...

  P = 10 W + 3·(10/3 W) = 20 W

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