Questions: A 22.0 ohm and 75.0 ohm resistor are in parallel, connected to a 5.00 V battery. How much current flows out of the battery? (Unit = A)

Solution Steps

To find the equivalent resistance \( R_{\text{eq}} \) of two resistors in parallel, we use the formula: \[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} \] where \( R_1 = 22.0 \, \Omega \) and \( R_2 = 75.0 \, \Omega \).

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{22.0} + \frac{1}{75.0} \]

Calculate the individual terms: \[ \frac{1}{22.0} \approx 0.0455 \quad \text{and} \quad \frac{1}{75.0} \approx 0.0133 \]

Add these values: \[ \frac{1}{R_{\text{eq}}} = 0.0455 + 0.0133 = 0.0588 \]

Now, take the reciprocal to find \( R_{\text{eq}} \): \[ R_{\text{eq}} = \frac{1}{0.0588} \approx 17.01 \, \Omega \]

Using Ohm's Law, \( V = IR \), we can find the total current \( I \) flowing out of the battery: \[ I = \frac{V}{R_{\text{eq}}} \] where \( V = 5.00 \, \text{V} \) and \( R_{\text{eq}} = 17.01 \, \Omega \).

\[ I = \frac{5.00}{17.01} \approx 0.2940 \, \text{A} \]

Final Answer