Figure 27-11a shows a multiloop circuit containing one ideal battery and four resistances with the following values:
`R_(1)=20 Omega, R_(2)=20 Omega, epsi=12V`,
`R_(3)=30 Omega, R_(4)=8.0 Omega`
(a) What is the current through the battery?
(b) What is the current `i_(2)` through `R_(2)?`
(c) What is the current `i_3" through "R_3?`

Explanation

Noting that the current through the battery must also be the current through `R_1` we see that we might find the current by applying the loop rule to a loop that includes `R_1` because the current would be included in the potential difference across `R_1`.
Incorrect method: Either the left-hand loop or the big loop should do. Noting that the emf arrow of the battery points upward, so the current the battery supplies is clockwisc, we might apply the loop rule to the left-hand loop, clockwise from point a. With i being the current through the battery, we would get
`+epsi-iR_(1)-iR_(2)-iR_(4)=0`
However, this equation is incorrect because it assumes that `R_1, R_2, and R_4`, all have the same current i. Resistances `R_1 and R_4` do have the same current, because the current passing through `R_4` must pass through the battery and then through `R_1` with no change in value. However, that current splits at junction point b- only part passes through `R_2` the rest through `R_3`.
Dead-end method: To distinguish the several currents in the circuit, we must label them individually as in Fig. 27-115. Then, circling clockwise from, we can write the loop rule for the teít-hand loop as
`+epsi+-i_(1)R_(1)-i_(2)R_(2)-i_(1)R_(4)=0`
Unfortunately, this equatinus contains two unknowns, `i_(1) and I_2` we would need at least one more equation find them.
Sucessful method : A much easier option is to simplify the circuit of Fig 27011b by finding equivalent series and thus cannot be replaced with an equivalent resistance. However `R_(2) and R_(3)` are not in series and thus cannot be replaced with an equivalent resistance. However `R_(2) and R_(3)` are in parallel, so we can use either Eq 27-24 or Eq 27-25 to find their equivalent resistance `R_(23)`. From the latter. `R_(23)=(R_(2)R_(3))/(R_(2)+R_(3))=((20 Omega)(30 Omega))/(50 Omega)=12 Omega`
We can now redraw the circuit as in Fig. 27 11c, note that the current through `R_(23)` must be `i_1` because charge that moves through `R_1 and R_4` must also move through `R_(23)` For this simple one-loop circuit, the loop rule (applied clockwise from point a as in Fig. 27-11d) yields `+epsi-i_(1)R_(1)-i_(1)R_(23)-i_(1)R_(4)=0`
Substituting the given data, we find
`12V-i_(1) (20 Omega) -i_(1) (12 Omega -i_(1) (8.0 Omega) =0`
which gives us `i_(1)=(12 V)/(40 Omega) =0.30 A`
(b) (1) We must now work backward from the equivalent circuit of Fig. 27-11d, where `R_(23)` has replaced `R_2 and R_3` (2) Because `R_2 and R_3` are in parallel, they both have the same potential difference across them as `R_(23).`
Working backward: We know that the current through `R_(23) is (1)=0.30 A.` Thus, we can use Eq. 26-8 (R= V/i) and Fig. 27-11e to find the potential difference `V_(23)" across "R_(23)`.
Setting `R_(23)= i_(1) R_(23) =(0.30 A) (12 Omega) =3.6V`
The potential difference across R is thus also 3.6 V (Fig. 27-11f), so the current `i_2" in "R_2` must be, by Eq. 26-8 and Fig 27-11g.
`i_(2)=(V_(2))/(R_(2))=(3.6V)/(20 Omega) =0.18A`
(c) We can answer by using either of techniques : (1) Apply Fiq 26-8 as we just did. (2) Use the junction rule, which tells us that at point b in Fig. 27-11b the incoming current `i_(1)` and the outgoing currents, `i_(2) and i_(3)` are related by `i_(1)=i_(2)+i_(3)`
Calculation : Rearranging this junction-rule result yields the result displayed in Fig 27-11g:
`i_(3)=i_(1)-i_(2)=0.30A-0.18A`
=0.12 A

Figure 27-11
(a) A circuit with an ideal battery. (b) Label the currents. (c) Replacing the parallel resistors with their equivalent (d)-(g) Working backward to find the currents through the paralel resistors.