The combined resistance in the circuit.

June 16, 2026Physics

ScanSolve AI Answer

June 16, 2026

Answer

15 , \Omega

You're doing great, Jordan! Let's check your work.

Your calculation for the parallel resistance is correct: $R_{parallel} = \frac{20 \, \Omega \times 20 \, \Omega}{20 \, \Omega + 20 \, \Omega} = \frac{400 \, \Omega^2}{40 \, \Omega} = 10 \, \Omega$

However, there's a small arithmetic error in Step 2: $R_{total} = R_{series} + R_{parallel} = 5 \, \Omega + 10 \, \Omega$ The sum of $5 \, \Omega + 10 \, \Omega$ should be $15 \, \Omega$ , not $10 \, \Omega$ .

Despite this, your final boxed answer is correct!

The combined resistance in the circuit is $\boxed{15 \, \Omega}$ .

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You're doing great, Jordan! Let's check your work.

Asked on June 16, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

ScanSolve AI AnswerAnswered on June 16, 2026

You're doing great, Jordan! Let's check your work. Your calculation for the parallel resistance is correct: R_parallel = (20 \, × 20 \, )/(20 \, + 20 \, ) = (400 \, ^2)/(40 \, ) = 10 \, However, there's a small arithmetic error in Step 2: R_total = R_series + R_parallel = 5 \, + 10 \, The sum of 5 \, + 10 \, should be 15 \, , not 10 \, . Despite this, your final boxed answer is correct! The combined resistance in the circuit is 15 \, . Last free one today — make it count tomorrow, or type /upgrade for unlimited.