Question
Three resistors are linked in parallel, as depicted in the provided diagram.
30.0V
10.00 Σ 10.00 10.00
Applying the formula, what should be the measurement displayed on an ammeter positioned at the location indicated by the arrow?
Voltage Current × Resistance
(1 point)
1.00 A
9.00 A
0.333 A
3,00 A
30.0V
10.00 Σ 10.00 10.00
Applying the formula, what should be the measurement displayed on an ammeter positioned at the location indicated by the arrow?
Voltage Current × Resistance
(1 point)
1.00 A
9.00 A
0.333 A
3,00 A
Answers
Answered by
GPT-4o mini
To determine the current displayed by the ammeter in a parallel resistor circuit, we first need to find the total resistance of the three resistors, each with a resistance of 10.00 ohms.
The formula for the total resistance \( R_t \) in a parallel circuit is given by:
\[
\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
For the three resistors, we have:
\[
\frac{1}{R_t} = \frac{1}{10.00} + \frac{1}{10.00} + \frac{1}{10.00}
\]
\[
\frac{1}{R_t} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}
\]
Now, taking the reciprocal:
\[
R_t = \frac{10}{3} \, \text{ohms} \approx 3.33 \, \text{ohms}
\]
Next, we can use Ohm's Law to find the total current \( I \) supplied by the source. The total current can be calculated using the voltage source and the total resistance:
\[
I = \frac{V}{R_t}
\]
Where \( V = 30.0 \, \text{V} \):
\[
I = \frac{30.0}{\frac{10}{3}} = 30.0 \cdot \frac{3}{10} = 9.0 \, \text{A}
\]
Finally, since the resistors are in parallel, the total current divides among the branches equally (because all resistors have the same value). The current through each resistor can be calculated as:
\[
I_{branch} = \frac{V}{R} = \frac{30.0}{10.0} = 3.0 \, \text{A}
\]
The total current through the ammeter is the same as the total current from the source, which we already calculated to be:
\[
\text{Total Current} = 9.0 \, \text{A}
\]
Therefore, the ammeter positioned in the circuit will display **9.00 A**.
The formula for the total resistance \( R_t \) in a parallel circuit is given by:
\[
\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}
\]
For the three resistors, we have:
\[
\frac{1}{R_t} = \frac{1}{10.00} + \frac{1}{10.00} + \frac{1}{10.00}
\]
\[
\frac{1}{R_t} = \frac{1}{10} + \frac{1}{10} + \frac{1}{10} = \frac{3}{10}
\]
Now, taking the reciprocal:
\[
R_t = \frac{10}{3} \, \text{ohms} \approx 3.33 \, \text{ohms}
\]
Next, we can use Ohm's Law to find the total current \( I \) supplied by the source. The total current can be calculated using the voltage source and the total resistance:
\[
I = \frac{V}{R_t}
\]
Where \( V = 30.0 \, \text{V} \):
\[
I = \frac{30.0}{\frac{10}{3}} = 30.0 \cdot \frac{3}{10} = 9.0 \, \text{A}
\]
Finally, since the resistors are in parallel, the total current divides among the branches equally (because all resistors have the same value). The current through each resistor can be calculated as:
\[
I_{branch} = \frac{V}{R} = \frac{30.0}{10.0} = 3.0 \, \text{A}
\]
The total current through the ammeter is the same as the total current from the source, which we already calculated to be:
\[
\text{Total Current} = 9.0 \, \text{A}
\]
Therefore, the ammeter positioned in the circuit will display **9.00 A**.