Calculate the current delivered by the battery, the emf (E) of the battery, and write down the resistance of N with a reason.

9.2.1 Current delivered by the battery Step 1: Identify the known values for the section of the circuit between points a and b. The effective resistance between points a and b is R_ab = 6 \, . The voltmeter V_1 reads the potential difference across points a and b, so V_ab = 5 \, V. Step 2: Calculate the total current (I_total) flowing through the circuit using Ohm's Law for the section between a and b. This current is the main current delivered by the battery. I_total = V_abR_ab I_total = (5 \, V)/(6 \, ) I_total ≈ 0.833 \, A The current delivered by the battery is 0.83 A. 9.2.2 Emf (E) of the battery Step 1: Calculate the total external resistance (R_ext) of the circuit. The external circuit consists of the effective resistance between a and b (R_ab) in series with resistor T (R_T). R_ext = R_ab + R_T R_ext = 6 \, + 1.5 \, R_ext = 7.5 \, Step 2: Calculate the EMF (E) of the battery using the formula E = I_total(R_ext + r), where r is the internal resistance. E = I_total(R_ext + r) E = ((5)/(6) \, A) (7.5 \, + 0.9 \, ) E = ((5)/(6)) (8.4 \, V) E = 7 \, V The EMF of the battery is 7 V. 9.2.3 Resistance of N The resistance of N is 3 . Reason: Resistors M and N are connected in series between points a and b. The total voltage across this series combination is V_ab = 5 \, V. Voltmeter V_2 measures the voltage across resistor M, which is V_M = 2.5 \, V. In a series circuit, the sum of individual voltages equals the total voltage, so V_N = V_ab - V_M = 5 \, V - 2.5 \, V = 2.5 \, V. Since the voltage across M (V_M = 2.5 \, V) is equal to the voltage across N (V_N = 2.5 \, V), and the current through series resistors is the same, their resistances must be equal (R_M = R_N). Given that the effective resistance of the series combination is R_ab = R_M + R_N = 6 \, , it follows that R_N = R_ab2 = (6 \, )/(2) = 3 \, . Send me the next one 📸