Question
11. A 15-volt power supply is used to power a circuit with two resistors in series. The voltage drop across R1
is 5 volts. What is the voltage drop across R2
?(1 point)
Responses
0 volts
0 volts
20 volts
20 volts
10 volts
10 volts
5 volts
5 volts
Question 12
12. A 10-volt power supply is used to power a circuit, producing a current of 2 A. If the total resistance of the circuit is doubled, how will this affect the circuit?(1 point)
Responses
The voltage drops across each resistor will all double to match the change in resistance.
The voltage drops across each resistor will all double to match the change in resistance.
The current will change so that the voltage drop for a complete loop through the circuit becomes 5 V.
The current will change so that the voltage drop for a complete loop through the circuit becomes 5 V.
The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
The voltage drops across each resistor will be unaffected because each will remain at 0 V.
The voltage drops across each resistor will be unaffected because each will remain at 0 V.
Question 13
13. A series circuit is formed with a 10-V power supply and two resistors, where R1
= 6 ohms and R2
= 4 ohms. The current through the resistors is I1
and I2
, respectively. What are the values of I1
and I2
?(1 point)
Responses
I1
= I2
= 1.0 A
I 1 = I 2 = 1.0 A
I1
= I2
= 0.5 A
I 1 = I 2 = 0.5 A
I1
= 1.67 A; I2
= 2.5 A
I 1 = 1.67 A; I 2 = 2.5 A
I1
= 0.6 A; I2
= 0.4 A
I 1 = 0.6 A; I 2 = 0.4 A
is 5 volts. What is the voltage drop across R2
?(1 point)
Responses
0 volts
0 volts
20 volts
20 volts
10 volts
10 volts
5 volts
5 volts
Question 12
12. A 10-volt power supply is used to power a circuit, producing a current of 2 A. If the total resistance of the circuit is doubled, how will this affect the circuit?(1 point)
Responses
The voltage drops across each resistor will all double to match the change in resistance.
The voltage drops across each resistor will all double to match the change in resistance.
The current will change so that the voltage drop for a complete loop through the circuit becomes 5 V.
The current will change so that the voltage drop for a complete loop through the circuit becomes 5 V.
The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
The voltage drops across each resistor will be unaffected because each will remain at 0 V.
The voltage drops across each resistor will be unaffected because each will remain at 0 V.
Question 13
13. A series circuit is formed with a 10-V power supply and two resistors, where R1
= 6 ohms and R2
= 4 ohms. The current through the resistors is I1
and I2
, respectively. What are the values of I1
and I2
?(1 point)
Responses
I1
= I2
= 1.0 A
I 1 = I 2 = 1.0 A
I1
= I2
= 0.5 A
I 1 = I 2 = 0.5 A
I1
= 1.67 A; I2
= 2.5 A
I 1 = 1.67 A; I 2 = 2.5 A
I1
= 0.6 A; I2
= 0.4 A
I 1 = 0.6 A; I 2 = 0.4 A
Answers
GPT-4o mini
Let's analyze each question one by one:
**Question 11:**
Given that the voltage across R1 is 5 volts in a series circuit with a total of 15 volts, we can find the voltage drop across R2.
The total voltage is the sum of the voltage drops across R1 and R2:
\[
V_{\text{total}} = V_{R1} + V_{R2}
\]
\[
15V = 5V + V_{R2}
\]
\[
V_{R2} = 15V - 5V = 10V
\]
**Answer:** 10 volts
---
**Question 12:**
You have a 10-volt power supply and a current of 2 A. Using Ohm's Law:
\[
V = I \times R
\]
To find the original resistance:
\[
R = \frac{V}{I} = \frac{10V}{2A} = 5 \, \text{ohms}
\]
If the total resistance is doubled, the new resistance \( R' \) will be:
\[
R' = 2 \times 5 \, \text{ohms} = 10 \, \text{ohms}
\]
Now, using Ohm's Law again to find the new current \( I' \) with the same voltage:
\[
I' = \frac{V}{R'} = \frac{10V}{10 \, \text{ohms}} = 1 \, \text{A}
\]
The overall voltage drop across the circuit remains 10V, but since the current decreased, the situation changes.
**Answer:** The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
---
**Question 13:**
Use Ohm's Law in a series circuit to find the current. The total resistance \( R_{total} \) is:
\[
R_{total} = R1 + R2 = 6 \, \text{ohms} + 4 \, \text{ohms} = 10 \, \text{ohms}
\]
Using the voltage supply of 10V:
\[
I = \frac{V}{R_{total}} = \frac{10V}{10 \, \text{ohms}} = 1 \, \text{A}
\]
Since the current is the same through all components in a series circuit:
\[
I1 = I2 = 1 \, \text{A}
\]
**Answer:** I1 = I2 = 1.0 A
---
To summarize:
- Question 11: 10 volts
- Question 12: The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
- Question 13: I1 = I2 = 1.0 A
**Question 11:**
Given that the voltage across R1 is 5 volts in a series circuit with a total of 15 volts, we can find the voltage drop across R2.
The total voltage is the sum of the voltage drops across R1 and R2:
\[
V_{\text{total}} = V_{R1} + V_{R2}
\]
\[
15V = 5V + V_{R2}
\]
\[
V_{R2} = 15V - 5V = 10V
\]
**Answer:** 10 volts
---
**Question 12:**
You have a 10-volt power supply and a current of 2 A. Using Ohm's Law:
\[
V = I \times R
\]
To find the original resistance:
\[
R = \frac{V}{I} = \frac{10V}{2A} = 5 \, \text{ohms}
\]
If the total resistance is doubled, the new resistance \( R' \) will be:
\[
R' = 2 \times 5 \, \text{ohms} = 10 \, \text{ohms}
\]
Now, using Ohm's Law again to find the new current \( I' \) with the same voltage:
\[
I' = \frac{V}{R'} = \frac{10V}{10 \, \text{ohms}} = 1 \, \text{A}
\]
The overall voltage drop across the circuit remains 10V, but since the current decreased, the situation changes.
**Answer:** The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
---
**Question 13:**
Use Ohm's Law in a series circuit to find the current. The total resistance \( R_{total} \) is:
\[
R_{total} = R1 + R2 = 6 \, \text{ohms} + 4 \, \text{ohms} = 10 \, \text{ohms}
\]
Using the voltage supply of 10V:
\[
I = \frac{V}{R_{total}} = \frac{10V}{10 \, \text{ohms}} = 1 \, \text{A}
\]
Since the current is the same through all components in a series circuit:
\[
I1 = I2 = 1 \, \text{A}
\]
**Answer:** I1 = I2 = 1.0 A
---
To summarize:
- Question 11: 10 volts
- Question 12: The current will change so that the voltage drop for a complete loop through the circuit remains at 10 V.
- Question 13: I1 = I2 = 1.0 A