A battery is connected to two resistors, A and B, with resistances 4 and 20 respectively. What is the current?

Step 1: Translate Question 3. The question asks: "List two methods to increase the strength of the magnetic field produced by a current-carrying circular coil."

Step 2: List the methods. The magnetic field strength of a current-carrying circular coil can be increased by:

• Increasing the current flowing through the coil.
• Increasing the number of turns in the coil.
• Inserting a soft iron core (ferromagnetic material) into the coil.
• Decreasing the radius of the coil.

We need to provide two methods. i) Increase the current flowing through the coil. ii) Increase the number of turns in the coil.

Step 3: Translate Question 4. The question asks: "How many times stronger is the magnetic field produced by a current-carrying circular coil with 5 turns compared to a coil with 1 turn and 2 turns?"

Step 4: Determine the relationship and calculate the ratios. The magnetic field strength ($B$) produced by a current-carrying circular coil is directly proportional to the number of turns ($N$) in the coil, assuming the current and radius are constant. $B \propto N$ • Comparing a 5-turn coil to a 1-turn coil: $\frac{B_{5 turns}}{B_{1 turn}} = \frac{5}{1} = 5$ The magnetic field is 5 times stronger than that of a 1-turn coil.

• Comparing a 5-turn coil to a 2-turn coil: $\frac{B_{5 turns}}{B_{2 turns}} = \frac{5}{2} = 2.5$ The magnetic field is 2.5 times stronger than that of a 2-turn coil.

Step 5: Translate Question 5. The question asks: "In the circuit diagram shown, which resistor, A or B, dissipates more electrical power? Give a reason."

Step 6: Analyze the circuit and calculate power. The resistors $R_A = 4 \, \Omega$ and $R_B = 20 \, \Omega$ are connected in series. In a series circuit, the current ($I$) flowing through each resistor is the same. The power ($P$) dissipated by a resistor is given by the formula $P = I^2 R$.

• Power dissipated by resistor A: $P_A = I^2 R_A = I^2 (4 \, \Omega)$ • Power dissipated by resistor B: $P_B = I^2 R_B = I^2 (20 \, \Omega)$

Step 7: Compare power dissipation and provide a reason. Since $R_B = 20 \, \Omega$ is greater than $R_A = 4 \, \Omega$, and the current $I$ is the same for both resistors, resistor B dissipates more power. $P_B > P_A$ Resistor B dissipates more electrical power. Reason: In a series circuit, the current flowing through all components is the same. The power dissipated by a resistor is directly proportional to its resistance ($P = I^2 R$). Since resistor B has a higher resistance ($20 \, \Omega$) than resistor A ($4 \, \Omega$), it dissipates more power.