Question

the following circuit draws 0.0212 amps of current from the cell. determine the current through the 110-ohm resistor, in amps.

Explanation:

Analyze the circuit topology

Using the Series and Parallel Circuits knowledge point

• The \(660\ \Omega\) resistor and the \(110\ \Omega\) resistor are connected in parallel.
• This parallel combination is in series with the \(330\ \Omega\) resistor and the cell.

Calculate the equivalent resistance of the parallel branch

Using the Equivalent Resistance knowledge point

$$ R_p = \frac{660 \times 110}{660 + 110} = \frac{72600}{770} = 94.545\ \Omega $$

Calculate the total equivalent resistance of the circuit

Using the Equivalent Resistance knowledge point

$$ R_{eq} = R_p + 330 = 94.545 + 330 = 424.545\ \Omega $$

Calculate the voltage across the parallel branch

Using the Ohm's Law knowledge point

• The total current from the cell is \(I_{total} = 0.0212\text{ A}\).
• The voltage across the parallel combination is:

$$ V_p = I_{total} \times R_p = 0.0212 \times 94.545 = 2.0044\text{ V} $$

Calculate the current through the 110-ohm resistor

Using the Ohm's Law knowledge point

• The current through the \(110\ \Omega\) resistor is:

$$ I_{110} = \frac{V_p}{110} = \frac{2.0044}{110} \approx 0.0182\text{ A} $$

Answer:

0.0182