Solve these questions---Question 12 (5 marks) [Circuit Diagram Description] Type: Electrical Circuit Diagram Main Elements: * Two points labeled A and B. * A voltage source labeled "10V". The positive terminal is on the left and the negative terminal is on the right, connected to the lower wire. * Three resistors are shown. * A 10Ω resistor is connected in series between point A and the next resistor. * A 25Ω resistor is connected in series after the 10Ω resistor. * After the 25Ω resistor, there is a parallel combination of two resistors: a 60Ω resistor and a 20Ω resistor. This parallel combination is connected in series with the 10Ω and 25Ω resistors, and the connection point after the parallel combination is point B. * The negative terminal of the 10V source is connected to the wire between the 25Ω resistor and the parallel combination. * The positive terminal of the 10V source is connected to the wire between point B and the parallel combination. For the circuit above, calculate: (a) The resistance between points A and B. [Handwritten Note]: R_eq = 10 + 25 + 1/(1/60 + 1/20) = 35 + 1/(4/60) = 35 + 60/4 = 35 + 15 = 50Ω. The handwritten calculation shows 35 + 1/(1/60 + 1/20) = 35 + 1/(4/60) = 35 + 15 = 50Ω. The handwritten final answer seems to be written as 40Ω initially and then corrected. (b) The potential difference across the 25Ω resistor. [Handwritten Note]: V=IR, I = V/R_eq. I = 10/40 (based on handwritten R_eq=40). I=1/4 A. V_25Ω = I * 25Ω = (1/4) * 25 = 25/4 = 6.25 V. The handwritten calculation shows V=IR...I=1/4...V=1/4 * 25...V=25/4...V=6.25. (c) The power dissipated by the 60Ω resistor. [Handwritten Note]: P=IV... Page Number: 9 Text at bottom: Toplogic - Physics [Part of Question 13 is visible on the right] Question 13 (5 m[...]) A circuit is made fro[m] 99 mΩ resistor to a [...] and has a radius of [...] capacity of 0.21 J g[^-1 ...] to its melting point. [...]

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Let's analyze this electrical circuit step by step. We have a 10V voltage source connected to a series combination of resistors. Starting from point A, we have a 10 ohm resistor in series with a 25 ohm resistor, followed by a parallel combination of 60 ohm and 20 ohm resistors, ending at point B. We need to find the equivalent resistance, voltage across the 25 ohm resistor, and power dissipated by the 60 ohm resistor. To find the equivalent resistance between points A and B, we first calculate the parallel combination of the 60 ohm and 20 ohm resistors. Using the parallel resistance formula, one over R parallel equals one over 60 plus one over 20, which equals 4 over 60, giving us 15 ohms. Then we add this to the series resistors: 10 ohms plus 25 ohms plus 15 ohms equals 50 ohms total resistance. For part b, we need to find the potential difference across the 25 ohm resistor. First, we calculate the total current using Ohm's law: I equals V over R, which is 10 volts divided by 50 ohms, giving us 0.2 amperes. Since the resistors are in series, the same current flows through each component. Using Ohm's law again, the voltage across the 25 ohm resistor is current times resistance: 0.2 amperes times 25 ohms equals 5 volts. For part c, we need to find the power dissipated by the 60 ohm resistor. First, we find the voltage across the parallel branch. Since the total current is 0.2 amperes and the parallel resistance is 15 ohms, the voltage is 0.2 times 15 equals 3 volts. This voltage appears across both parallel resistors. The current through the 60 ohm resistor is 3 volts divided by 60 ohms, which equals 0.05 amperes. Finally, the power dissipated is V squared over R: 3 squared over 60 equals 9 over 60, which equals 0.15 watts. Let's summarize our complete solution to this circuit analysis problem. For part a, we found the equivalent resistance between points A and B to be 50 ohms by calculating the parallel combination of 60 and 20 ohm resistors as 15 ohms, then adding the series resistances. For part b, the potential difference across the 25 ohm resistor is 5 volts, calculated using the total current of 0.2 amperes. For part c, the power dissipated by the 60 ohm resistor is 0.15 watts, found using the voltage across the parallel branch. This problem demonstrates the fundamental principles of circuit analysis using Ohm's law and series-parallel resistance calculations.