Question
Three 10 Ω resistors are connected in parallel to a 12-V battery. What is the total current in the circuit? (Hint: use your response from the previous question)
*
1 point
3.6 A
.4 A
1.3 A
*
1 point
3.6 A
.4 A
1.3 A
Answers
GPT-4o mini
To find the total current in the circuit with three 10 Ω resistors connected in parallel to a 12-V battery, we first need to calculate the equivalent resistance (R_eq) of the parallel resistors.
The formula for the equivalent resistance of \( n \) resistors in parallel is given by:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
For three resistors, all having the same resistance (10 Ω):
\[
\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}
\]
So,
\[
R_{eq} = \frac{10}{3} \, \Omega \approx 3.33 \, \Omega
\]
Next, we use Ohm's law (V = IR) to find the total current (I) in the circuit. Rearranging the equation gives:
\[
I = \frac{V}{R_{eq}}
\]
Inserting the values:
\[
I = \frac{12V}{\frac{10}{3} \, \Omega} = \frac{12 \times 3}{10} = \frac{36}{10} = 3.6 \, A
\]
So the total current in the circuit is **3.6 A.**
The formula for the equivalent resistance of \( n \) resistors in parallel is given by:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
For three resistors, all having the same resistance (10 Ω):
\[
\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}
\]
So,
\[
R_{eq} = \frac{10}{3} \, \Omega \approx 3.33 \, \Omega
\]
Next, we use Ohm's law (V = IR) to find the total current (I) in the circuit. Rearranging the equation gives:
\[
I = \frac{V}{R_{eq}}
\]
Inserting the values:
\[
I = \frac{12V}{\frac{10}{3} \, \Omega} = \frac{12 \times 3}{10} = \frac{36}{10} = 3.6 \, A
\]
So the total current in the circuit is **3.6 A.**