Combining Resistors in Series & Parallel Video Lessons

Concept

Problem: In the circuit of Fig. E26.15, each resistor represents a light bulb. Let R1 = R2 = R3 = 4.50 Ω and ε = 9.00 V. (a) What is the current in each of the remaining bulbs R1, R2, and R3?(b) What is the power dissipated in each of the remaining bulbs?

FREE Expert Solution

Equivalent resistance for 2 resistors in parallel:

$\overline{){{\mathbf{R}}}_{{\mathbf{eq}}}{\mathbf{=}}\frac{{\mathbf{R}}_{\mathbf{1}}{\mathbf{R}}_{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{1}}\mathbf{+}{\mathbf{R}}_{\mathbf{2}}}}$

Equivalent resistance for resistors in series:

$\overline{){{\mathbf{R}}}_{{\mathbf{eq}}}{\mathbf{=}}{{\mathbf{R}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{2}}}{\mathbf{+}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{+}}{{\mathbf{R}}}_{{\mathbf{n}}}}$

Current:

$\overline{){\mathbf{i}}{\mathbf{=}}\frac{\mathbf{V}}{\mathbf{R}}}$

Power:

$\overline{){\mathbf{P}}{\mathbf{=}}{{\mathbf{i}}}^{{\mathbf{2}}}{\mathbf{R}}}$

(a)

Req = R + (R)(R)/(R + R) = R + R/2 = 1.5R

Req = (1.5)(4.50) = 6.75Ω

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Problem Details

In the circuit of Fig. E26.15, each resistor represents a light bulb. Let R1 = R2 = R3 = 4.50 Ω and ε = 9.00 V.

(a) What is the current in each of the remaining bulbs R1, R2, and R3?

(b) What is the power dissipated in each of the remaining bulbs?