Question

1.a battery of 12v connected is parallel connected with resistor of 6 ohm, 6 ohm, and 5 ohm
A.eguivalent resistance
B.current through the battery
Find those questions?

Answer 1

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 \(I \approx 6.4 \, A\)