Pregunta
\( \frac { ( 56 k \Omega ) ( 12 k \Omega ) } { 56 k \Omega + 12 k \Omega } = \)
Ask by Wheeler Chavez. in Pakistan
Mar 01,2025
Solución de inteligencia artificial de Upstudy
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The equivalent resistance is approximately 9.88 kΩ.
Solución
To find the equivalent resistance for two resistors in parallel, use the formula:
R_eq = (R₁ × R₂) / (R₁ + R₂)
Here, R₁ = 56 kΩ and R₂ = 12 kΩ.
Step 1: Multiply the resistances:
56 kΩ × 12 kΩ = 672 (kΩ)²
Step 2: Add the resistances:
56 kΩ + 12 kΩ = 68 kΩ
Step 3: Divide the product by the sum:
R_eq = 672 (kΩ)² / 68 kΩ = 672 / 68 kΩ
Simplify the fraction:
Divide numerator and denominator by 2: (672 ÷ 2) / (68 ÷ 2) = 336/34 kΩ
Divide again by 2: (336 ÷ 2) / (34 ÷ 2) = 168/17 kΩ
Therefore, the equivalent resistance is 168/17 kΩ, which is approximately 9.88 kΩ.
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Simplifique esta solución The Deep Dive
To find the equivalent resistance \( R \) of two resistors in parallel, we use the formula: \[ R = \frac{R_1 \times R_2}{R_1 + R_2} \] Here, \( R_1 = 56 k\Omega \) and \( R_2 = 12 k\Omega \). Plugging in these values: \[ R = \frac{(56 k\Omega)(12 k\Omega)}{56 k\Omega + 12 k\Omega} = \frac{(672 k\Omega^2)}{68 k\Omega} \] Now calculating the right side: \[ R = \frac{672}{68} k\Omega \approx 9.882 k\Omega \] So, the equivalent resistance of a \( 56 k\Omega \) resistor in parallel with a \( 12 k\Omega \) resistor is approximately \( 9.88 k\Omega \).
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