Question

In a parallel circuit three resistors(4 ohm's, 6 ohm's and 10 ohm's), calculate the total resistance if the current in each branch if a 12 Voltage battery is connected

Answer 1

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:

1. For \( R_1 = 4 \, \Omega \):
\[
I_1 = \frac{12 \, V}{4 \, \Omega} = 3 \, A
\]

2. For \( R_2 = 6 \, \Omega \):
\[
I_2 = \frac{12 \, V}{6 \, \Omega} = 2 \, A
\]

3. 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): \( I_3 = 1.2 \, A \)