The document describes solving a circuit problem to determine the emf of an ideal battery. It provides the current through one resistor and all resistor values. It then calculates potential differences across individual resistors based on Ohm's law and uses Kirchhoff's laws to relate the currents and voltages throughout the circuit in a step-by-step manner. Finally, it determines that the emf of the battery is 58.7 Volts.

1. Chapter 27, Problem 033 In the figure the current in resistance 6 is i6 = 1.49 A and the
resistances are R1 8.30 ?, and R6 = 4.69 ?, what is the emf of the ideal battery? R2 = R3 = 233 ?,
R4 14.2 ?, R5 Ri R2 so 16 Number Units the tolerance is +/-9%
Solution
V 6 = potential difference across R 6 = i 6 R 6 = (1.49)(4.69) = 6.9881 Volts
Since R 5 and R 6 are in series
i 5 = i 6 = 1.49 A
V 5 = potential difference across R 5 = i 5 R 5 = (1.49)(8.3) = 12.367 Volts
combination of R 5 and R 6 is in parallel with R 4 , hence
V 4 = V 5 + V 6
V 4 = 12.367 + 6.9881 = 19.3551 Volts
i 4 = current in R 4 = V 4 /R 4 = 19.3551/14.2 = 1.363 A
Also,
i 2 = i 4 + i 5
i 2 = (1.363) + 1.49 = 2.853 A
V 2 = Voltage across R 2 = i 2 R 2 = (2.853) (2.33) = 6.65 Volts
V 3 = V 2 + V 4 = 6.65 + 19.355 = 26 Volts
i 3 = V 3 /R 3 = 26/2.33 = 11.2 A
i 1 = i 2 + i 3 = 2.853 + 11.2 = 14.05 A

2. V 1 = i 1 R 1 = (14.05) (2.33) = 32.7 Volts
E = emf of the battery = V 1 + V 3 = 32.7 + 26 = 58.7 Volts