Question

Look at the following circuit

22 V

22 Ohms

8 Ohms

10 Ohms

What is the TOTAL current that will be drawn from the energy source?

(1 point)

1.31 A

0.29 A

0.55 A

0.84 A

Answer 1

To determine the total current drawn from the energy source, we first need to analyze how the resistances are configured in the circuit. Let's assume a common setup: 22 Ohms in series with a parallel combination of 8 Ohms and 10 Ohms.

1. First, we calculate the equivalent resistance of the 8 Ohm and 10 Ohm resistors in parallel.

The formula for equivalent resistance \( R_{eq} \) of two resistors in parallel is:

\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
\]

Substituting the values:

\[
\frac{1}{R_{eq}} = \frac{1}{8} + \frac{1}{10}
\]

Finding a common denominator (40):

\[
\frac{1}{R_{eq}} = \frac{5}{40} + \frac{4}{40} = \frac{9}{40}
\]

Now, taking the reciprocal:

\[
R_{eq} = \frac{40}{9} \approx 4.44 \, \text{Ohms}
\]

2. Now combine this equivalent resistance with the 22 Ohm resistor in series:

\[
R_{total} = R_{22} + R_{eq} = 22 + \frac{40}{9}
\]

To simplify this, convert 22 Ohms to a fraction with a denominator of 9:

\[
22 = \frac{198}{9}
\]
Thus,

\[
R_{total} = \frac{198}{9} + \frac{40}{9} = \frac{238}{9} \approx 26.44 \, \text{Ohms}
\]

3. Finally, we use Ohm’s Law to find the total current \( I \):

\[
I = \frac{V}{R_{total}} = \frac{22 \, \text{V}}{\frac{238}{9} \, \text{Ohms}} = \frac{22 \times 9}{238}
\]

Calculating this gives:

\[
I = \frac{198}{238} \approx 0.831 \, \text{A}
\]

So, the total current drawn from the energy source is approximately **0.84 A**.

The answer is **0.84 A**.