Current Electricity · Ammeter Reading

Reading of the Ammeter (12 V Network)

A 12 V source drives the redrawn network. Find the equivalent resistance and the current read by the ammeter A.

Top 4 Ω, bottom 4 Ω, and the middle (two 8‖8 blocks in series) all span the same two nodes.

Given

EMF: 12 V (ideal source).
Top branch: 4 Ω; bottom branch: 4 Ω.
Middle: left block 8‖8 and right block 8‖8, in series.

Asked

Equivalent resistance R_eq and the ammeter current I.

Formula

Parallel \( \tfrac{1}{R}=\sum\tfrac{1}{R_i} \); series adds; Ohm's law:

\[ I = \frac{V}{R_{eq}} \]

Solution

STEP 1 — Collapse each 8‖8 block

\[ 8 \parallel 8 = \frac{8\times8}{16} = 4\ \Omega \quad(\text{each block}) \]

STEP 2 — The two blocks are in series (middle path)

\( 4 + 4 = 8\ \Omega \).

STEP 3 — Three parallel branches: 4 Ω, 8 Ω, 4 Ω

\[ \frac{1}{R_{eq}} = \frac{1}{4}+\frac{1}{8}+\frac{1}{4} = \frac{2+1+2}{8} = \frac{5}{8} \;\Rightarrow\; R_{eq} = \frac{8}{5} = 1.6\ \Omega \]

STEP 4 — Ohm's law for the ammeter (total) current

\[ I = \frac{V}{R_{eq}} = \frac{12}{1.6} = 7.5\ \text{A} \]

Ammeter Reading R_eq = 1.6 Ω, I = 7.5 A

Short Cut

Two equal resistors in parallel halve. Each 8‖8 is instantly 4 Ω; the middle path is 4 + 4 = 8 Ω. For the final 4 ‖ 8 ‖ 4: pair the equals first (4 ‖ 4 = 2), then 2 ‖ 8 = 1.6 Ω.

The ammeter sits in the main line (in series with the source), so it reads the total current, not a branch current: 12 / 1.6 = 7.5 A.

Given → Asked → Formula → Solution → Short Cut · R_eq = 8/5 Ω, I = 7.5 A