Question

aining Time: 2 hours, 21 minutes, 49 seconds. astion Completion Status: QUESTION 24 Consider the circuit below and answer the following questions. In this sketch, V=8V, RI-200, R2-300, and R3-4000 ww a) What is the equivalent resistance of the circuit? b) How much current is drawn from the battery? How much power is dissipated by R3? Please write your answers in the space below and email your work TTTT Paragraph Arial - 3120 • E- E - T-... XD09 T' T. O S. - Chek Save and submit to send the Save Allowers SUS esc स 17 B ord N + 2 s % 3 & 4 5 6 7 8 ON WE “า Louis T lo Yi

Answer 1

So here are in parallel, and $R_{1}$ is in series to this parallel configuration. So the equivalent resistence is, $R_{eq} = R_{1} + \left(\frac1{R_{2}} + \frac1{R_{3}}\right)^{-1} = 200 + \left( \frac1{300} + \frac1{400} \right)^{-1} ~ \Omega = 371.429 ~ \Omega.$

Using Ohm's Law, $V = I \cdot R_{eq} \implies I = \frac{V}{R_{eq}}.$

Here, $V = 8 ~ V, R_{eq}= 371.429 ~ \Omega.$ Hence, $I = \frac{8}{371.429} ~ A = 0.0215385 ~ A.$

Now for a parallel circuit, the voltage remains constant. Using this information we can compute $I_3$ the current through $R_3$ . Now $I_1 = I \implies V_1 = 4.30769 ~ V.$ So the voltage drop across
R2, R3
is So $I_2 = \frac{3.69231}{R_2} = \frac{3.6923}{300} ~ A = 0.0123077 ~ A.$

Now the power is given by, $P_{2} = I_{2}^2 \cdot R_{2} = ( 0.0123077 )^2 \times 300 ~ W = 0.0454438 ~ W.$

Note: The power can also be found out via, $P_{2} = \frac { V_{2}^2 } { R_{2} } = \frac { ( 3.69231 ) ^ 2 }{ 300 } ~ W = 0.0454438 ~ W.$