Problem — two resistors in parallel

Problem

A $12\ \text{V}$ battery with negligible internal resistance powers two resistors in parallel: $R_1 = 8{,}0\ \Omega$ and $R_2 = 24\ \Omega$. Calculate the total current delivered and the p.d. across the resistors.

Solution

P.d. across the resistors. In a parallel combination on an ideal battery (negligible internal resistance) each branch sees the full battery voltage: $V_R = 12\ \text{V}$

Equivalent resistance of the parallel combination: $R_\text{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{8{,}0 \cdot 24}{8{,}0 + 24} = \frac{192}{32} = 6{,}0\ \Omega$

Total current delivered: $I_\text{tot} = \frac{V}{R_\text{eq}} = \frac{12}{6{,}0} = 2{,}0\ \text{A}$ Check by branch: $I_1 = 12/8{,}0 = 1{,}5\ \text{A}$, $I_2 = 12/24 = 0{,}5\ \text{A}$, sum $= 2{,}0\ \text{A}$ ✓.

$\ev{I_\text{tot} = 2{,}0\ \text{A} \qquad V_R = 12\ \text{V}}$

Links

Topics: Circuiti elettrici Concepts: Resistenze in serie e parallelo