In a parallel circuit, the total (or equivalent) resistance \( R_t \) can be calculated using the formula:
\[ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
where \( R_1, R_2, \) and \( R_3 \) are the resistances of the individual resistors.
Given the resistances:
- \( R_1 = 4 , \Omega \)
- \( R_2 = 6 , \Omega \)
- \( R_3 = 10 , \Omega \)
Substituting the values into the equation:
\[ \frac{1}{R_t} = \frac{1}{4} + \frac{1}{6} + \frac{1}{10} \]
To find a common denominator, we can use 60 (the least common multiple of 4, 6, and 10):
\[ \frac{1}{R_t} = \frac{15}{60} + \frac{10}{60} + \frac{6}{60} = \frac{15 + 10 + 6}{60} = \frac{31}{60} \]
Now, we can find \( R_t \):
\[ R_t = \frac{60}{31} \approx 1.935 , \Omega \]
Now we need to calculate the current through each resistor using Ohm's Law:
\[ I = \frac{V}{R} \]
For each resistor:
-
For \( R_1 = 4 , \Omega \): \[ I_1 = \frac{12 , V}{4 , \Omega} = 3 , A \]
-
For \( R_2 = 6 , \Omega \): \[ I_2 = \frac{12 , V}{6 , \Omega} = 2 , A \]
-
For \( R_3 = 10 , \Omega \): \[ I_3 = \frac{12 , V}{10 , \Omega} = 1.2 , A \]
Summary:
- Total resistance \( R_t \approx 1.935 , \Omega \)
- Current through \( R_1 \) (4 Ω): \( I_1 = 3 , A \)
- Current through \( R_2 \) (6 Ω): \( I_2 = 2 , A \)
- Current through \( R_3 \) (10 Ω): \( I_3 = 1.2 , A \)