Question

In this circuit the battery provides 3 V, the resistance R1 is 7 Ω, and R2 is 5 Ω. What is the current through resistor R2? Give your answer in units of Amps. An Amp is 1 Coulomb of charge flowing through a cross-sectional area of the wire per second - that's a lot of charge per second and will warm up a typical wire quite a bit! Most devices have circuits with larger resistors - kLaTeX: \OmegaΩ (103 LaTeX: \OmegaΩ) and MLaTeX: \OmegaΩ (106 LaTeX: \OmegaΩ) are common.

Answer 1

Answer:

The current pass the $R_2$ is  $I = 0.25 A$

Explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

    The voltage  is  $V = 3V$

     The first resistance is  $R_1 = 7 \Omega$

     The second resistance is  $R_2 = 5 \Omega$

Since the resistors are connected in series their equivalent resistance is  

       $R_{eq} = R_1 +R_2$

Substituting values

         $R_{eq} = 7 + 5$

         $R_{eq} = 12 \Omega$

Since the resistance are connected in serie the current passing through the circuit  is the same current passing through $R_2$ which is mathematically evaluated as

        $I = \frac{V}{R_{eq}}$

Substituting values  

      $I = \frac{3}{12}$

      $I = 0.25 A$