To solve the given circuit questions, let's analyze the circuit step by step.
A. Equivalent Resistance
In a parallel circuit, the formula for finding the equivalent resistance (\(R_{eq}\)) of multiple resistors is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
Where:
- \(R_1 = 6 , \Omega\)
- \(R_2 = 6 , \Omega\)
- \(R_3 = 5 , \Omega\)
Now, substituting the values:
\[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{5} \]
Calculating each term: \[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{5} = \frac{1}{6} + \frac{1}{6} + \frac{1}{5} \]
Finding a common denominator, which is 30: \[ \frac{1}{6} = \frac{5}{30}, \quad \frac{1}{6} = \frac{5}{30}, \quad \frac{1}{5} = \frac{6}{30} \]
So: \[ \frac{1}{R_{eq}} = \frac{5}{30} + \frac{5}{30} + \frac{6}{30} = \frac{16}{30} \]
Taking the reciprocal of \( \frac{16}{30} \) to find \( R_{eq} \): \[ R_{eq} = \frac{30}{16} = 1.875 , \Omega \]
B. Current Through the Battery
To find the total current (\(I\)) coming from the battery, we can use Ohm's Law, which states:
\[ I = \frac{V}{R_{eq}} \]
Where:
- \(V = 12 , V\)
- \(R_{eq} = 1.875 , \Omega\)
Now substituting the values:
\[ I = \frac{12}{1.875} \approx 6.4 , A \]
Summary of the Answers:
A. The equivalent resistance \(R_{eq} \approx 1.875 , \Omega\)
B. The current through the battery \(I \approx 6.4 , A\)