Given:

• $R_1 = 2$ Ω, $R_2 = 4$ Ω, $R_3 = 6$ Ω (series)
• $V = 12$ V

Part (a): Equivalent resistance

Series: Resistances add $R_{eq} = R_1 + R_2 + R_3 = 2 + 4 + 6 = 12 \text{ Ω}$

Part (b): Total current

$I = \frac{V}{R_{eq}} = \frac{12}{12} = 1.0 \text{ A}$

Part (c): Voltage across each

Series: Same current through each $V_1 = IR_1 = (1.0)(2) = 2.0 \text{ V}$

Check: $V_1 + V_2 + V_3 = 2 + 4 + 6 = 12$ V ✓

Answer:

• (a) R_eq = 12 Ω
• (b) I = 1.0 A
• (c) V₁ = 2.0 V, V₂ = 4.0 V, V₃ = 6.0 V

Given:

• $R_1 = 6$ Ω, $R_2 = 3$ Ω, $R_3 = 2$ Ω (parallel)
• $V = 12$ V

Part (a): Equivalent resistance

Parallel: Reciprocal resistances add $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Part (b): Total current

$I_{total} = \frac{V}{R_{eq}} = \frac{12}{1.0} = 12 \text{ A}$

Part (c): Current through each

Parallel: Same voltage across each $I_1 = \frac{V}{R_1} = \frac{12}{6} = 2.0 \text{ A}$

Check: $I_1 + I_2 + I_3 = 2 + 4 + 6 = 12$ A ✓

Answer:

• (a) R_eq = 1.0 Ω (less than any individual!)
• (b) I_total = 12 A
• (c) I₁ = 2.0 A, I₂ = 4.0 A, I₃ = 6.0 A

Given:

• $R_1 = 6$ Ω, $R_2 = 3$ Ω, $R_3 = 2$ Ω (parallel)
• $V = 12$ V

Part (a): Equivalent resistance

Parallel: Reciprocal resistances add $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

Part (b): Total current

$I_{total} = \frac{V}{R_{eq}} = \frac{12}{1.0} = 12 \text{ A}$

Part (c): Current through each

Parallel: Same voltage across each $I_1 = \frac{V}{R_1} = \frac{12}{6} = 2.0 \text{ A}$

Check: $I_1 + I_2 + I_3 = 2 + 4 + 6 = 12$ A ✓

Answer:

• (a) R_eq = 1.0 Ω (less than any individual!)
• (b) I_total = 12 A
• (c) I₁ = 2.0 A, I₂ = 4.0 A, I₃ = 6.0 A

Solution:

Series connection: Same current through all resistors

(a) Equivalent resistance: R_eq = R₁ + R₂ + R₃ = 2.0 + 4.0 + 6.0 = 12 Ω

(b) Current: I = V/R_eq = 24/12 = 2.0 A (same through all)

(c) Voltage across each: V₁ = IR₁ = 2.0 × 2.0 = 4.0 V V₂ = IR₂ = 2.0 × 4.0 = 8.0 V V₃ = IR₃ = 2.0 × 6.0 = 12 V

Check: 4.0 + 8.0 + 12 = 24 V ✓

(d) Power in R₂: P₂ = I²R₂ = (2.0)²(4.0) = 16 W

Given:

• $R_1 = 4$ Ω, $R_2 = 8$ Ω (series)
• $R_3 = 6$ Ω (parallel with series combo)
• $V = 12$ V

Solution:

Step 1: Find equivalent of series combination. $R_{series} = R_1 + R_2 = 4 + 8 = 12 \text{ Ω}$

Step 2: This 12 Ω is in parallel with 6 Ω. $\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R_3}$

Or using product-over-sum: $R_{eq} = \frac{(12)(6)}{12 + 6} = \frac{72}{18} = 4.0 \text{ Ω}$

Step 3: Total current from battery. $I_{total} = \frac{V}{R_{eq}} = \frac{12}{4.0} = 3.0 \text{ A}$

Answer: I_total = 3.0 A

Solution:

Parallel connection: Same voltage across all resistors (12 V)

(a) Equivalent resistance: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ 1/R_eq = 1/6.0 + 1/3.0 + 1/2.0 = 1/6 + 2/6 + 3/6 = 6/6 = 1 R_eq = 1.0 Ω

(b) Current through each: I₁ = V/R₁ = 12/6.0 = 2.0 A I₂ = V/R₂ = 12/3.0 = 4.0 A I₃ = V/R₃ = 12/2.0 = 6.0 A

(c) Total current: I_total = I₁ + I₂ + I₃ = 2.0 + 4.0 + 6.0 = 12 A

Or: I_total = V/R_eq = 12/1.0 = 12 A ✓

(d) Total power: P_total = VI_total = 12 × 12 = 144 W

Or: P = V²/R_eq = 144/1.0 = 144 W ✓

Given:

• $R_1 = 4$ Ω, $R_2 = 8$ Ω (series)
• $R_3 = 6$ Ω (parallel with series combo)
• $V = 12$ V

Solution:

Step 1: Find equivalent of series combination. $R_{series} = R_1 + R_2 = 4 + 8 = 12 \text{ Ω}$

Step 2: This 12 Ω is in parallel with 6 Ω. $\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R_3}$

Or using product-over-sum: $R_{eq} = \frac{(12)(6)}{12 + 6} = \frac{72}{18} = 4.0 \text{ Ω}$

Step 3: Total current from battery. $I_{total} = \frac{V}{R_{eq}} = \frac{12}{4.0} = 3.0 \text{ A}$

Answer: I_total = 3.0 A