Question 1:

The correct answer regarding the resistance of an ideal voltmeter and an ideal ammeter is:

B. Voltmeter resistance: infinite; Ammeter resistance: zero.

• Voltmeter: Ideally, it should have infinite resistance to prevent it from drawing any current from the circuit, ensuring accurate voltage measurements.
• Ammeter: Ideally, it should have zero resistance to avoid affecting the current in the circuit it measures.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 2:

To find the charge flowing around the circuit per second, we can use Ohm's Law and the relationship between current (I), charge (Q), and time (t):

Ohm's Law states:

I= ε /R
​
where I is the current, ε is the electromotive force (emf), and R is the resistance of the load.

The current (I) is also defined as the charge per unit time:

I= Q/t

Since we are interested in the charge flowing per second (which corresponds to the current), we can simplify this to:

Q=I×t

t=1:

so Q=I

Substituting Ohm's Law into this gives:

Q = ε /R

Conclusion
Thus, the charge flowing around the circuit per second is Q = ε /R, answer B.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 3:

To find the reading on the voltmeter in the potential divider circuit, we can use the formula for the voltage across a resistor in a series circuit:

V_x = (R_x/R_total) x V_suppy

Given V_suppy =12V, R_x = 100kΩ and the total resistance R_total = 150kΩ

V_x = (R_x/R_total) x V_suppy
​
Substituting in the values:

V_x = (100kΩ/150kΩ) x 12 ​
​
V_x = (2/3)×12=8V

Conclusion
The reading on the voltmeter, which is across the 100kΩ resistor, is 8 V, which is answer D

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 4:

When the voltmeter is replaced with a cheap voltmeter that has a resistance of 100 kΩ, it forms a parallel circuit with the 100 kΩ resistor. This changes the effective resistance of the circuit.

Step 1: Calculate the equivalent resistance of the 100 kΩ resistor in parallel with the 100 kΩ voltmeter.
The formula for resistors in parallel is:

1/R_total = 1/R₁ + 1/R₂ + 1/R₃ ......

Substitute the values, in kΩ:

1/R_total = 1/100 + 1/100 = 2/100 = 1/50

R_total =50 (kΩ)

Step 2: Calculate the total resistance in the circuit.
Now the total resistance in the circuit is:

=50kΩ+50kΩ=100kΩ

Step 3: Calculate the current in the circuit.
Using Ohm's Law:

I = V/R = 12 / 100x10³

​I =0.00012A=120μA

Step 4: Calculate the voltage across the 100 kΩ resistor (now in parallel).
The voltage across the 100 kΩ resistor (or the voltmeter) can be calculated using:

V =I×R

=120μA×50kΩ = (120x10^-6) x (50 x10³)
​
or =0.00012A×50000Ω= 6V

Conclusion
The new reading on the voltmeter is C. 6 V.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 5:

The properties of resistivity and charge carrier density in copper cables can be understood as follows:

1. Resistivity: This is a material property that is inherent to the material itself. For copper, resistivity is generally low due to its excellent conductivity.

2. Charge Carrier Density: In metals like copper, the charge carrier density (number of free electrons available for conduction) is typically high.

Comparison of Cables P and Q:

• Both cables are made of copper, so their resistivity is the same and low.
• The diameter of cable P is smaller than that of Q, which means for the same length, cable P has a higher resistance due to its smaller cross-sectional area. However, this does not change the intrinsic properties of the material.

Conclusion

Therefore, the correct answer regarding the resistivity and charge carrier density in the copper cables is:

C. Resistivity: low; Charge carrier density: high.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 6:

To determine the resistance of rod Q, we can use the formula for resistance based on the dimensions of the rod:

R= ρL/A
​
where:
R is the resistance,
ρ is the resistivity (a material property),
L is the length of the rod,
A is the cross-sectional area.

Given:
Rod P has a resistance of R.
Rod Q has double the radius of rod P.

Step 1: Calculate the area of the cross-section
The cross-sectional area A of a rod is given by:

A=πr²

If the radius of rod P is r, then the radius of rod Q is 2r.

The area of rod Q will be:
​
A_Q=π(2r)² = 4πr² = 4A_P (4 x the area of rod P)
​

Step 2: Calculate the resistance of rod Q
Assuming both rods have the same length L, the resistance of rod Q can be expressed as:

R_Q = ρL/(4A) = = ¼.ρL/A = ¼ R_P
​
Conclusion
Thus, the resistance of rod Q is answer A. ¼ R_P

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 7:

To find the energy lost as heat in the cable, we can use the formula for electrical power and the relationship between energy, power, and time.

Step 1: Calculate the current (I)
Using Ohm's Law:

I= V/R

where V is the EMF of the cell (3 V) and R is the resistance of the rod.

So:

I= 3/R
​

Step 2: Calculate the power (P):
The power dissipated as heat in the resistor is given by:

P=I² R

Substituting the expression for current:

P=(3/R)² .R = 9/R
​

Step 3: Calculate the energy lost as heat (E)
Energy lost as heat in a time t is given by:

E=P×t

Substituting the expression for power:

E= 9/R . t

or E = 9t/R
​
Conclusion
Thus, the energy lost as heat in the cable in time t is 9t/R which is answer B.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 8:

To analyze the circuit with a 12 V cell, an internal resistor r, and an external resistor R=2Ω, we can use Kirchhoff's voltage law.

Step 1: Understand the circuit

The total EMF from the cell is 12V. The current I flows through the circuit.
The voltage drop across the internal resistor is V = I.r.
The voltage drop across the external resistor V = I.2.

Step 2: Apply Kirchhoff's Voltage Law
According to Kirchhoff's law, the sum of the voltage drops in the circuit equals the EMF provided by the cell:

EMF = Voltage drop across r_internal + Voltage drop across R_external

This gives:

12=I.2+I.r

Step 3: Factor the equation
We can factor out I:

12=I(2+r)

Conclusion
Thus, the correct formula for the circuit is: C: 12=I(2+r)

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 9:

To find the reading on the voltmeter, which is across the external resistor R=2Ω, we can use Ohm's Law. The voltage V across the resistor is given by:

V=I.R

Given:
Current I=4A and resistance R=2Ω:

Step 1: Calculate the voltage across the external resistor
Substituting the values into the formula:

V=4A.2Ω=8V

Conclusion
Thus, the reading on the voltmeter is:

D. 8 V.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.

Question 10:

To find the value of the internal resistance r in the circuit, we can use the information we have:

The total EMF is 12V. The external resistance R=2Ω. The current I=4A.

The voltage drop across the external resistor V=I.R is 8V.

Step 1: Calculate the voltage drop across the internal resistor
The total voltage drop in the circuit is given by:

EMF = Voltage drop across r_internal + Voltage drop across R_external

We can express this as:

12=V+I.r

Substituting the known values:

12=8+4r

Step 2: Solve for r:
Rearranging the equation:

12−8=4r
4 = 4r, ​
r =1Ω

Conclusion:
Thus, the value of r is answer B. 1 Ω.

*These A.I. responses have been individually checked to ensure they match the accepted answer, but explanations may still be incorrect. Responses may give guidance but the A.I. might not be able to answer the question! This is particularly the case for questions based on diagrams, which the A.I. typically cannot interpret.
Grade Gorilla uses Gemini, Deepseek and a range of other A.I. chatbots to generate the saved responses. Some answers have had human intervention for clarity or where the A.I. has not been able to answer the question.