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VARIABLE EQUATIONS & INEQUALITIES UNIT TEST: THE ELECTRIC CURRENT, /(AMPS), PASSING THROUGH AN ELECTRONIC COMPONENT VARIES DIRE...Question
VARIABLE EQUATIONS & INEQUALITIES UNIT TEST:
THE ELECTRIC CURRENT, /(AMPS), PASSING THROUGH AN ELECTRONIC COMPONENT VARIES DIRECTLY WITH THE APPLIED VOLTAGE, V (VOLTS), ACCORDING TO THE RELATIONSHIP V= IR WHERE R, IS THE RESISTANCE MEASURED IN OHMS (Ω).
WONG IS TRYING TO FIND OUT THE RESISTANCE, R, OF AN UNKNOWN COMPONENT. HE LABELS THE RESISTANCE OF THE UNKNOWN COMPONENT AS xΩ. THE RESISTANCE, R, OF A CIRCUIT CONTAINING THESE UNKNOWN COMPONENTS IS (5 X + 20) Ω. A 90-VOLT POTENTIAL DIFFERENCE ACROSS THE CIRCUIT PRODUCES A CURRENT OF 2.25 AMPS.
WHAT IS THE RESISTANCE OF THE UNKNOWN COMPONENT? EXPLAIN HOW YOU ARRIVED AT YOUR ANSWER. IN 1 OR 2 SHORT SENTENCES.
THE ELECTRIC CURRENT, /(AMPS), PASSING THROUGH AN ELECTRONIC COMPONENT VARIES DIRECTLY WITH THE APPLIED VOLTAGE, V (VOLTS), ACCORDING TO THE RELATIONSHIP V= IR WHERE R, IS THE RESISTANCE MEASURED IN OHMS (Ω).
WONG IS TRYING TO FIND OUT THE RESISTANCE, R, OF AN UNKNOWN COMPONENT. HE LABELS THE RESISTANCE OF THE UNKNOWN COMPONENT AS xΩ. THE RESISTANCE, R, OF A CIRCUIT CONTAINING THESE UNKNOWN COMPONENTS IS (5 X + 20) Ω. A 90-VOLT POTENTIAL DIFFERENCE ACROSS THE CIRCUIT PRODUCES A CURRENT OF 2.25 AMPS.
WHAT IS THE RESISTANCE OF THE UNKNOWN COMPONENT? EXPLAIN HOW YOU ARRIVED AT YOUR ANSWER. IN 1 OR 2 SHORT SENTENCES.
Answers
GPT-4o mini
To find the resistance of the unknown component, we first use Ohm's Law, which states \( V = IR \). Given the voltage \( V = 90 \) volts and current \( I = 2.25 \) amps, we calculate the total resistance \( R \) as follows:
\[
R = \frac{V}{I} = \frac{90}{2.25} = 40 \, \text{Ω}.
\]
Next, we equate this to the total resistance equation \( 5x + 20 = 40 \) and solve for \( x \):
\[
5x + 20 = 40 \implies 5x = 20 \implies x = 4 \, \text{Ω}.
\]
Thus, the resistance of the unknown component is \( 4 \, \text{Ω} \).
\[
R = \frac{V}{I} = \frac{90}{2.25} = 40 \, \text{Ω}.
\]
Next, we equate this to the total resistance equation \( 5x + 20 = 40 \) and solve for \( x \):
\[
5x + 20 = 40 \implies 5x = 20 \implies x = 4 \, \text{Ω}.
\]
Thus, the resistance of the unknown component is \( 4 \, \text{Ω} \).