Question

A 10 $\Omega$ resistor is connected in series with a parallel combination of two resistors (4 $\Omega$ and 6 $\Omega$). The circuit is powered by a 24 V battery with an internal resistance of 1 $\Omega$. Calculate the current through the 6 $\Omega$ resistor. $\frac{i_1}{}=\frac{R_2}{}$

Answer 1

0.72 A

Answer 2

Solution:

1. Find Equivalent Resistance of the Parallel Combination:

   $$R_{\text{parallel}} = \frac{4 \times 6}{4+6} = \frac{24}{10} = 2.4\,\Omega$$

2. Determine Total Circuit Resistance:

   Including the 1 Ω internal resistance and 10 Ω resistor in series:

   $$R_{\text{total}} = 1 + 10 + 2.4 = 13.4\,\Omega$$

3. Calculate Total Current from the Battery:

   $$I_{\text{total}} = \frac{24\,V}{13.4\,\Omega} \approx 1.79\,A$$

4. Voltage Across the Parallel Combination:

   $$V_{\text{parallel}} = I_{\text{total}} \times R_{\text{parallel}} \approx 1.79\,A \times 2.4\,\Omega \approx 4.30\,V$$

5. Current Through the 6 Ω Resistor Using Ohm’s Law:

   $$I_{6} = \frac{V_{\text{parallel}}}{6\,\Omega} \approx \frac{4.30\,V}{6\,\Omega} \approx 0.72\,A$$

   Alternatively, using current division:

   $$I_{6} = I_{\text{total}} \times \frac{R_{\text{other}}}{4+6} = 1.79\,A \times \frac{4}{10} \approx 0.72\,A$$