Question

In the circuit shown below, $V_5 = 100 V$, $R_1 = 30 \Omega$, $R_2 = 60 \Omega$, $R_3 = 90 \Omega$, $R_4 = 45 \Omega$, and $R_5 = 30 \Omega$. The current flowing through resistor $R_3$ is \(\underline{\hspace{1cm}}\) A. (rounded off to two decimal places)

• A. +0.30
• B. +0.21
• C. -0.21
• D. -0.30

Hint:

Use the current divider rule when dealing with resistors in parallel and series to calculate the current through specific resistors.

Answer 1

The Correct Option is B

First, we will calculate the total resistance in the circuit. The resistors are connected in a combination of series and parallel. Let's break down the calculation step by step: 1. Combine $R_1$ and $R_2$ in parallel: The formula for two resistors in parallel is: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \] Substituting values: \[ \frac{1}{R_{\text{parallel}}} = \frac{1}{30} + \frac{1}{60} = \frac{2}{60} + \frac{1}{60} = \frac{3}{60} = \frac{1}{20} \] Thus, \[ R_{\text{parallel}} = 20 \, \Omega \] 2. Add $R_3$ in series with $R_{\text{parallel}$:} \[ R_{\text{total1}} = R_{\text{parallel}} + R_3 = 20 + 90 = 110 \, \Omega \] 3. Combine $R_4$ and $R_5$ in parallel: \[ \frac{1}{R_{\text{parallel2}}} = \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{45} + \frac{1}{30} \] \[ \frac{1}{R_{\text{parallel2}}} = \frac{2}{90} + \frac{3}{90} = \frac{5}{90} \] Thus, \[ R_{\text{parallel2}} = \frac{90}{5} = 18 \, \Omega \] 4. Add $R_{\text{parallel2}$ in series with $R_{\text{total1}}$:} \[ R_{\text{total}} = R_{\text{total1}} + R_{\text{parallel2}} = 110 + 18 = 128 \, \Omega \] 5. Calculate the total current using Ohm's Law: \[ I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{100}{128} = 0.78125 \, \text{A} \] 6. Finally, calculate the current through $R_3$: Since $R_3$ is in the path with $R_{\text{parallel}}$, we use the current divider rule. The current through $R_3$ is: \[ I_{R_3} = I_{\text{total}} \times \frac{R_{\text{parallel}}}{R_{\text{total}}} = 0.78125 \times \frac{20}{128} = 0.21 \, \text{A} \] Thus, the current through $R_3$ is +0.21 A, corresponding to option (B).