Two resistors, 20 $\Omega$ and 30 $\Omega$, are connected in parallel. A third resistor, 24 $\Omega$, is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 24- $\Omega$ resistor is

Question

Two resistors, 20 $\Omega$ and 30  $\Omega$, are connected in parallel. A third resistor, 24  $\Omega$, is connected in series with the parallel combination. A 36-V power source is connected to the series-parallel combination. The current through the 24- $\Omega$ resistor is

Quizplus · Accepted Answer

First, calculate the equivalent resistance of the two parallel resistors: $R_{eq} = \frac{1}{\frac{1}{20} + \frac{1}{30}} = 12 \Omega$. Then, add the series resistor to find the total resistance: $R_{total} = R_{eq} + 24 = 36 \Omega$. Finally, use Ohm's law ($V = IR$) to find the current: $I = \frac{V}{R_{total}} = \frac{36V}{36\Omega} = 1.0 A$.

Two Resistors, 20 Ω\OmegaΩ And 30 Ω\OmegaΩ , Are Connected in Parallel Ω\OmegaΩ

Two Resistors, 20 $\Omega$ And 30 $\Omega$ , Are Connected in Parallel $\Omega$