In (Figure 1), [tex]R1 = 3.00 \, \Omega[/tex], [tex]R2 = 8.00 \, \Omega[/tex], and [tex]R3 = 3.00 \, \Omega[/tex]. The battery has negligible internal resistance. The current [tex]i2[/tex] through [tex]R2[/tex] is [tex]4.00 \, A[/tex].

Question

13-06-2025
Physics

High School

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In (Figure 1), [tex]R1 = 3.00 \, \Omega[/tex], [tex]R2 = 8.00 \, \Omega[/tex], and [tex]R3 = 3.00 \, \Omega[/tex]. The battery has negligible internal resistance. The current [tex]i2[/tex] through [tex]R2[/tex] is [tex]4.00 \, A[/tex].

Answer :

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SumanaPaul SumanaPaul · Answer 1 · 2024-03-16T14:52:29-04:00

Final answer:

The question involves using Ohm's law and principles of series and parallel circuits to calculate the currents through resistors in an electric circuit.

Explanation:

The student's question pertains to electric circuits and specifically deals with calculating currents through different resistors when connected in series and parallel configurations. To solve these problems, one must apply Ohm's law and the principles of series and parallel circuits.

The current through R2 is given, and from there, we can use Ohm's law (V=IR) to find the voltage across R2 and R3. Since R1 is in series with the parallel combination of R2 and R3, the voltage across R1 can be calculated by the voltage supplied by the battery minus the voltage across the parallel combination of R2 and R3. Then, applying Ohm's law again, we can determine the current through R1. Knowing that the total current from the battery is the sum of currents through the series and parallel paths, we can ensure the consistency of our calculations by equating the power supplied by the battery to the total power dissipated by the resistors.