Question
Three resistors are linked in parallel, as illustrated in the diagram provided.
A voltage of 30.0 V is connected to three resistors, each 10.0 Ohms. There is an arrow pointing next to the third resistor.
What reading should the ammeter display when positioned at the location indicated by the arrow?
(1 point)
Responses
9.00 A
9.00 A
1.00 A
1.00 A
0.333 A
0.333 A
3.00 A
A voltage of 30.0 V is connected to three resistors, each 10.0 Ohms. There is an arrow pointing next to the third resistor.
What reading should the ammeter display when positioned at the location indicated by the arrow?
(1 point)
Responses
9.00 A
9.00 A
1.00 A
1.00 A
0.333 A
0.333 A
3.00 A
Answers
Answered by
GPT-4o mini
To determine the current reading on the ammeter positioned next to the third resistor in a parallel circuit with three resistors of 10.0 Ohms each, we start by calculating the total current flowing into the parallel combination of resistors.
1. **Calculate the equivalent resistance (R_eq)** for the three resistors in parallel using the formula:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
where \( R_1 = R_2 = R_3 = 10.0 \, \Omega \).
Thus,
\[
\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}
\]
Therefore,
\[
R_{eq} = \frac{10}{3} \approx 3.33 \, \Omega
\]
2. **Calculate the total current (I_total)** coming from the voltage source using Ohm's law:
\[
I_{total} = \frac{V}{R_{eq}} = \frac{30.0 \, V}{3.33 \, \Omega} \approx 9.00 \, A
\]
3. **Determine the current through each resistor**:
Since the resistors are in parallel, the voltage across each resistor is the same (30.0 V). The current through an individual resistor can be calculated using Ohm's law:
\[
I_n = \frac{V}{R_n} = \frac{30.0 \, V}{10.0 \, \Omega} = 3.00 \, A
\]
4. **Determine the current through the third resistor**:
Since each resistor experiences the same voltage and has the same resistance, the current through the third resistor (or any of the resistors) would be 3.00 A.
Therefore, the reading on the ammeter positioned next to the third resistor should be **3.00 A**.
1. **Calculate the equivalent resistance (R_eq)** for the three resistors in parallel using the formula:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
where \( R_1 = R_2 = R_3 = 10.0 \, \Omega \).
Thus,
\[
\frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}
\]
Therefore,
\[
R_{eq} = \frac{10}{3} \approx 3.33 \, \Omega
\]
2. **Calculate the total current (I_total)** coming from the voltage source using Ohm's law:
\[
I_{total} = \frac{V}{R_{eq}} = \frac{30.0 \, V}{3.33 \, \Omega} \approx 9.00 \, A
\]
3. **Determine the current through each resistor**:
Since the resistors are in parallel, the voltage across each resistor is the same (30.0 V). The current through an individual resistor can be calculated using Ohm's law:
\[
I_n = \frac{V}{R_n} = \frac{30.0 \, V}{10.0 \, \Omega} = 3.00 \, A
\]
4. **Determine the current through the third resistor**:
Since each resistor experiences the same voltage and has the same resistance, the current through the third resistor (or any of the resistors) would be 3.00 A.
Therefore, the reading on the ammeter positioned next to the third resistor should be **3.00 A**.